MCQ 511 Mark
The value of $(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$ is
Answer$(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$
$=(1+\text{i})(1-1)(1-\text{i})(1+1) \ \big(\because\text{i}^2=-1, \beta=-\text{i and} \ \text{i}^4=1\big)$
$=(1+\text{i})(0)(1-\text{i})(2)$
$=0$
View full question & answer→MCQ 521 Mark
The values of $k$ for which the quadratic equation $k x^2+1=k x+3 x-11 x^2$ has real and equal roots are:
- A
$-11, -3$
- B
$5, 7$
- ✓
$5, -7$
- D
AnswerCorrect option: C. $5, -7$
The given equation is $k x^2+1=k x+3 x-11 x^2$ which can be written as.
$ k x^2+11 x^2-k x-3 x+1= $
$ \Rightarrow(k+11) x^2-(k+3) x+1=0$
For equal and real roots, the discriminant of $(k+11) x^2-(k+3) x+1=0$.
$ \therefore(k+3)^2-4(k+11)=0 $
$ \Rightarrow k^2+2 k-35=0 $
$\Rightarrow(k-5)(k+7)=0 $
$ \Rightarrow k=5,-7$
Hence, the equation has real and equal roots when $k=5,-7$.
View full question & answer→MCQ 531 Mark
If $\text{z}=\frac{-2}{1+\sqrt{3}},$ then the value of $\text{arg (z)}$ is:
- A
$\pi$
- B
$\frac{\pi}{3}$
- ✓
$\frac{2\pi}{3}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{2\pi}{3}$
$\text{z}=\frac{-2}{1+\sqrt{3}}$
Rationalising $z$, we get,
$\text{z}=\frac{-2}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{-2+\text{i}2\sqrt{3}}{1+3}$
$\Rightarrow\text{z}=\frac{-1+\text{i}\sqrt{3}}{2}$
$\Rightarrow\text{z}=\frac{-1}{2}+\frac{\text{i}\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Since, $z$ lies in the second quadrant.
Therefore, $\text{arg(z)}=\pi-\frac{\pi}{3}$
$=\frac{2\pi}{3}$
View full question & answer→MCQ 541 Mark
$6i$ is point on $...........?$
- A
$x -$ axis
- ✓
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect option: B. $y -$ axis
Since real part of complex number is zero.
So, it is plotted on imaginary axis
i.e. $y -$ axis.
$6i$ is point on $y -$ axis.
View full question & answer→MCQ 551 Mark
If the complex number $\text{z}=\text{x}+\text{iy}$ satisfies the condition $|\text{z}+1|=1,$ then $z$ lies on:
AnswerCorrect option: B. circle with centre $(-1, 0)$ and radius $1$
$|\text{z}+1|=1$
$\Rightarrow|\text{z+1|}^2=1^2$
$\Rightarrow(\text{z}+1)\overline{(\text{z}+1)}=1$
$\Rightarrow(\text{z}+1)(\overline{z}+1)=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}+1=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
Since, $\text{z}=\text{x}+\text{iy}$
$\therefore\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
$\Rightarrow(\text{x}+\text{iy})(\text{x}-\text{iy})+\text{x}+\text{iy}+\text{x}- \text{iy}=0$
$\Rightarrow\text{x}^2+\text{y}^2+\text{2x}=0$
$\Rightarrow(\text{x}+1)^2+(\text{y}-0)^2=1^2$
which is the equation of a circle with the center $(-1,\ 0)$ and radius $1.$
View full question & answer→MCQ 561 Mark
Choose the correct answer. If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $z = 1 + 2i,$ then $|f(z)|$ is:
- ✓
$\frac{|\text{z}|}{2}$
- B
$|\text{z}|$
- C
$2|\text{z}|$
- D
AnswerCorrect option: A. $\frac{|\text{z}|}{2}$
Given that, $\text{z}=1+2\text{i}$
$|\text{z}|=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
Now, $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1-2\text{i}}{1-1-4\text{i}^2-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}$
$=\frac{3-\text{i}}{2-2\text{i}}$
$=\frac{3-\text{i}}{2-2\text{i}}\times\frac{2+2\text{i}}{2+2\text{i}}$
$=\frac{6+6\text{i}-2\text{i}-2\text{i}^2}{4-4\text{i}}$
$=\frac{6+4\text{i}+2}{4+4}$
$=\frac{8+4\text{i}}{8}$
$=1+\frac{1}{2}\text{i}$
So, $|\text{f(z)}|=\sqrt{(1)+\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt{5}}{2}$
$=\frac{|\text{z}|}{2}$
View full question & answer→MCQ 571 Mark
Let $S$ denotes the set of all real values of the parameter $'a\ '$ for which every solution of the inequality $\log\frac{1}{2} \text{x}^2\geq \log\frac{1}{2} (\text{x} + 2)$ is the solution of the inequality $49\text{x}^2-4\text{a}^4\leq0$ What is the value of $S?$
- A
$(-\infty,-\sqrt{7})\cup(\sqrt{7},\infty)$
- ✓
$(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
- C
$(-\sqrt{7, }\sqrt{7})$
- D
$\Big[-\sqrt{7, }\sqrt{7}\Big]$
AnswerCorrect option: B. $(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
We have, $\log\frac{1}{2} \text{x}^2\geq \log\frac{1}{2} (\text{x} + 2)$
$\Rightarrow\text{x}^2\leq\text{x}+2$
$\Rightarrow-1\leq\text{x}\leq2$
And, $49\text{x}^2-4\text{a}^4\leq0$
$\text{ i.}\text{e.}\text{x}^2\leq\frac{4\text{a}^4}{49}$
$\Rightarrow\frac{-2\text{a}^2}{7}\leq\text{a}\leq\frac{2\text{a}^2}{7}$
From the above equations,
$\frac{-2\text{a}^2}{7}\leq-1$ and $2\leq\frac{2\text{a}^2}{7}$
$\text{i.}\text{e.} a^2 \in\frac{7}{2} $ and $\text{a}^2\geq7$
$\Rightarrow\text{a}\in(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
So, $\text{S}=(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
View full question & answer→MCQ 581 Mark
If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $\text{z}=1+2\text{i},$ then $|\text{f(z)}|$ is:
- ✓
$\frac{|\text{z}|}{2}$
- B
$|\text{z}|$
- C
$2|\text{z}|$
- D
AnswerCorrect option: A. $\frac{|\text{z}|}{2}$
$\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1+2\text{i}}{1-(1^2+2^2\text{i}^2+4\text{i})}$
$=\frac{6-2\text{i}}{1-1+4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}\times\frac{4+4\text{i}}{4+4\text{i}}$
$=\frac{24+24\text{i}-8\text{i}-8\text{i}^2}{4^2-4^2\text{i}^2}$
$=\frac{24+16\text{i}+8}{16+16}$
$=\frac{32+16\text{i}}{32}$
$=1+\frac{1}{2}\text{i}$
since $\text{z}=1+2\text{i},$
$\because|\text{z}|=\sqrt{(1)^2+(2)^2}$
$=\sqrt{1+4}$
$=\sqrt{5}$
$\therefore|\text{f}\text{(z)|}=\sqrt{(1)^1+(\frac{1}{2})^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt5}{2}$
$=\frac{|\text{z}|}{2}$
View full question & answer→MCQ 591 Mark
$a + i b > c + id$ can be explained only when:
- A
$b = 0, c = 0$
- ✓
$b = 0, d = 0$
- C
$a = 0, c = 0$
- D
$a = 0, d = 0$
AnswerCorrect option: B. $b = 0, d = 0$
View full question & answer→MCQ 601 Mark
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2.5.10.17...(1+\text{n}^2)=$
- A
$\text{a}-\text{ib}$
- B
$\text{a}^2-\text{b}^2$
- ✓
$\text{a}^2+\text{b}^2$
- D
AnswerCorrect option: C. $\text{a}^2+\text{b}^2$
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get,
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$
View full question & answer→MCQ 611 Mark
Choose the correct answer.
The point represented by the complex number $2 - i$ is rotated about origin through an angle $\frac{\pi}{2}$ in the clockwise direction, the new position of point is:
- A
$1 + 2i$
- ✓
$-1 - 2i$
- C
$2 + i$
- D
$-1 + 2i$
AnswerCorrect option: B. $-1 - 2i$
Given that, $\text{z}=2-\text{i}$
If $z$ rotated through an angle of $\frac{\pi}{2}$ about the origin in clockwise direction.
Then the new position $=\text{z}\cdot\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\Big[\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)\Big]$
$=(2-\text{i})(0-\text{i})$
$=-1-2\text{i}$
View full question & answer→MCQ 621 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$ have roots $\alpha$ and $\beta$.Then, what will be the value of $\sin\alpha + \sin\beta$
AnswerCorrect option: A. $\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
Given, $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$
So, this implies $\text{a}\cos\theta=\text{c}-\text{b}\sin\theta$
Now squaring both the sides we get,
$(\text{a}\cos\theta)^2=(\text{c}-\text{b}\sin\theta)^2$
$\text{a}^2\cos^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2(1-\sin^2\theta)=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2-\text{a}^2\sin^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
Now rearranging the elements,
$\text{a}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta+(\text{c}^2-\text{a}^2)=\theta$
So, as sum of the roots are in the form $\frac{-\text{b}}{\text{a}}$ if there is a quadratic equation $\text{a}\text{x}^2=\text{bx}+\text{c}=0$
Now, we can conclude that
$\sin\alpha+\sin\beta=\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
View full question & answer→MCQ 631 Mark
The value of $p$ and $q (\text{P}\neq0,\ \text{q}\neq0)$ for which $p, q$ are the roots of the equation $x^2 + px + q = 0$ are:
- ✓
$p = 1, q = −2$
- B
$p = −1, q = −2$
- C
$p = −1, q = 2$
- D
$p = 1, q = 2$
AnswerCorrect option: A. $p = 1, q = −2$
It is given that, $p$ and $q (\text{P}\neq0,\ \text{q}\neq0)$are the roots of the equation $x^2 + px + q = 0$
$\therefore$ Sum of roots $= p + q = −p$
$\Rightarrow 2p + q = 0 ...(1)$
Product of roots $= pq = q$
$\Rightarrow q (p − 1) = 0$
$\Rightarrow p = 1, q = 0$
Now, substituting $p = 1$ in $(1),$ we get,
$2 + q = 0$
$\Rightarrow q = −2$
View full question & answer→MCQ 641 Mark
Value of $i ($iota$)$ is:
- A
$-1$
- B
$1$
- ✓
$(-1)^\frac{1}{2}$
- D
$(-1)^\frac{1}{4}$
AnswerCorrect option: C. $(-1)^\frac{1}{2}$
Iota is used to denote complex number.
The value of $i ($iota$)$ is $\sqrt{-1}\text{ i.e.}(-1)^\frac{1}{2}$
View full question & answer→MCQ 651 Mark
Find mirror image of point representing $x + i y$ on real axis:
- A
$(x, y)$
- B
$(-x, -y)$
- C
$(-x, y)$
- ✓
$(x, -y)$
AnswerCorrect option: D. $(x, -y)$
Mirror image of point $(x, y)$ on real axis is $(x, -y).$
Since real axis is acting as mirror $x-$coordinate remains same whereas $y-$coordinate gets inverted.
So, $(x, -y)$ is mirror image of $(x, y)$ on real axis.
View full question & answer→MCQ 661 Mark
If $\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib},$ then $\text{a}^2+\text{b}^2=$
Answer$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$\Big|\frac{1-\text{ix}}{1+\text{ix}}\Big|=\big|\text{a}+\text{ib}\big|$
$\Rightarrow\frac{\sqrt{1^2+\text{x}^2}}{\sqrt{1^2+\text{x}^2}}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=1$
Squaring both the sides, we get:
$\text{a}^2+\text{b}^2=1$
View full question & answer→MCQ 671 Mark
The set of all values of $m$ for which both the roots of the equation $x^2− (m + 1) x + m + 4 = 0$ are real and negative, is:
AnswerCorrect option: C. $[-4, -3]$
The roots of the quadratic equation $x^2− (m + 1) x + m + 4 = 0$ will be real, if its discriminant is greater than or equal to zero.
$\therefore (m + 1)^2 - 4 (m + 4) > 0$
$\Rightarrow (m - 5) (m + 3) > 0$
$\Rightarrow m < -3$ or $m > 5 ...(1)$
It is also given that, the roots of $x^2− (m + 1) x + m + 4 = 0$ are negative.
So, the sum of the roots will be negative.
$\therefore$ Sum of the roots $< 0$
$\Rightarrow m + 1 < 0 ....(2)$
and product of zeros $> 0$
$\Rightarrow m + 4 > 0 ...(3)$
From $(1), (2)$ and $(3),$ we get,
$[−4, −3]$
View full question & answer→MCQ 681 Mark
In $z = 4 + i,$ what is the real part?
AnswerIn $z = a + b i\, a$ is real part and $b$ is imaginary part.
So, in $4 + i,$ real part is $4.$
View full question & answer→MCQ 691 Mark
According to De Moivre’s theorem what is the value of $\text{z}^\frac{1}{\text{n}}$
- A
$\text{r}^\frac{1}{\text{n}}\big[\cos2\text{kn}+\theta)+\text{i}\sin(2\text{kn}+\theta)\big]$
- B
$\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}-\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
- ✓
$\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}+\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
- D
$\text{r}^\frac{1}{\text{n}}\big[\cos2\text{kn}+\theta)-\text{i}\sin(2\text{kn}+\theta)\big]$
AnswerCorrect option: C. $\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}+\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
If $n$ is any integer, then $(\cos\theta+\text{i}\sin\theta)^\text{n}=\cos(\text{n}\theta)+\text{i}\sin(\text{n}\theta)$
Writing the binomial expansion of $(\cos\theta+\text{i}\sin\theta)^\text{n}$ and equating real parts of $\cos(\text{n}\theta)$ and the imaginary part to $\sin(\text{n}\theta)$ we ge
$\cos\text{n}\theta=\cos^\text{n}\theta-\text{n}\text{c}_2\cos^\text{n}-2\theta\sin^2\theta+\text{nc}_4\cos^\text{n-4}\theta\sin^4\theta+\dots\\\sin(\text{n}\theta)=\text{nc}_1\cos^\text{n-1}\theta\sin\theta-\text{nc}_3\cos^\text{n-3}\theta\sin^3\theta+\dots$
If, $n$ is a rational number, then one of the value of $(\cos\theta+\text{i}\sin\theta)^\text{n}=\cos(\text{n}\theta)+\text{i}\sin(\text{n}\theta)$
If, $\text{n}=\frac{\text{p}}{\text{q}},$
where, $p$ and $q$ are integers $(\text{q}>\theta)$ and $p, q$ have no common factor, then $(\cos\theta+ \text{i}\sinθ)^\text{n}$ has $q$ distinct values one of which is $(\cos\theta+ \text{i}\sinθ)^\text{n}$
If, $\text{z}^\frac{1}{\text{n}}\bigg[\frac{\cos(2\text{k}\pi+\theta)}{\text{n}}\bigg]+\bigg[\frac{\text{i}\sin(2\text{k}\pi+\theta)}{\text{n}}\bigg]$where $k = 0, 1, 2, ……….., n – 1.$
View full question & answer→MCQ 701 Mark
A real number $'a\ '$ is called a good number if the inequality$\frac{(2\text{x}^2 - 2\text{x} -3)}{(\text{x}^2 + \text{x} + 1)}\leq a$ is satisfied for all real $x.$ What is the set of all real numbers?
- A
$\Big(\infty,\frac{10}{3}\Big]$
- B
$\Big(\frac{10}{3 },\infty\Big)$
- ✓
$\Big[\frac{10}{3},\infty\Big)$
- D
$\Big[\frac{-10}{3},\infty\Big)$
AnswerCorrect option: C. $\Big[\frac{10}{3},\infty\Big)$
We have, $\frac{(2\text{x}^2 - 2\text{x} -3)}{(\text{x}^2 + \text{x} + 1)}\leq\text{a}\forall\text{x}\in\text{R}$
$\Rightarrow2\text{x}^2 – 2\text{x} – 3\leq \text{a}(\text{x}^2 + \text{x} + 1)\forall \text{x} \in \text{R}$
$\Rightarrow(2 – \text{a})\text{x}^2 – (2 – \text{a})\text{x} – (3 – \text{a})\forall \text{ x} \in\text{R}$
$2-\text{a}<0$ and $(2 – \text{a})\text{x}^2 – 4(2 – \text{a})(3 – \text{a})\leq 0 \forall \text{ x } \in \text{R}$
So, $\text{a}>2$ and $\text{a}\leq2$ or $\text{a}\geq\frac{10}{3}$
$\Rightarrow\text{a}\geq\frac{10}{3}$
$\therefore\text{a}\in\Big[\frac{10}{3},\infty\Big)$
View full question & answer→MCQ 711 Mark
The sum of two complex numbers $a + ib$ and $c + id$ is a real number if
- A
$a + c = 0$
- ✓
$b + d = 0$
- C
$a + b = 0$
- D
$b + c = 0$
AnswerCorrect option: B. $b + d = 0$
It is given that
$z_1=a+i b$ and
$z_2=c+i d$
Then
$z_1+z_2=(a+c)+i(b+d)$
Now
$\left(z_1+z_2\right)$ is purely real.
Then the imaginary part has to be $0.$
Hence
$b + d = 0.$
View full question & answer→MCQ 721 Mark
Let $z_1$ and $_2$ be two com plex num bers with α and $\beta$ as their principal ar gu ments such that $\alpha+\beta>\pi,$ then principal arg $(z_1z_2)$ is given by:
- A
$\alpha+\beta+\pi$
- ✓
$\alpha+\beta-\pi$
- C
$\alpha+\beta+2\pi$
- D
$\alpha+\beta$
AnswerCorrect option: B. $\alpha+\beta-\pi$
View full question & answer→MCQ 731 Mark
Choose the correct answer. If $z$ is a complex number, then:
- A
$|\text{z}^2|>|\text{z}|^2$
- ✓
$|\text{z}^2|=|\text{z}|^2$
- C
$|\text{z}^2|<|\text{z}|^2$
- D
$|\text{z}^2|\geq|\text{z}|^2$
AnswerCorrect option: B. $|\text{z}^2|=|\text{z}|^2$
Let $z=x+y i$
$|z|=|x+y i|$ and $|z|^2=|x+y i|^2$
$\Rightarrow|z|^2=x^2+y^2 \ldots \text { (i) }$
Now, $z^2=x^2+y^2 i^2+2 x y i$
$z^2=x^2-y^2+2 x y i$
$|\text{z}^2|=\sqrt{(\text{x}^2-\text{y}^2)^2+(2\text{xy})^2}$
$=\sqrt{\text{x}^4+\text{y}^4-2\text{x}^2\text{y}^2+4\text{x}^2\text{y}^2}$
$=\sqrt{\text{x}^4+\text{y}^4+2\text{x}^2\text{y}^2}$
$=\sqrt{(\text{x}^2+\text{y}^2)^2}$
So, $|\text{z}|^2=\text{x}^2+\text{y}^2=|\text{z}|^2$
So, $|\text{z}|^2=|\text{z}^2|$
View full question & answer→MCQ 741 Mark
Find mirror image of point representing $x + i y$ on imaginary axis:
- A
$(x, y)$
- B
$(-x, -y)$
- ✓
$(-x, y)$
- D
$(x, -y)$
AnswerCorrect option: C. $(-x, y)$
Mirror image of point $(x, y)$ on imaginary axis is $(-x, y).$
Since imaginary axis is acting as mirror $y-$coordinate remains same whereas $x -$ coordinate gets inverted.
So, $(-x, y)$ is mirror image of $(x, y)$ on imaginary axis.
View full question & answer→MCQ 751 Mark
If $\text{i}^2=-1,$ then the sum $\text{i}+\text{i}^2+\text{i}^3+...$ upto $1000$ terms is equal to:
Answer$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i and} \ \text{i}^4=1]$
$=\text{i}-1-\text{i}+1$
$=0$
Similarly, the sum of the next four terms of the series will be equal to $0.$
This is because the powers of $i$ follow a cyclicity of $4.$
Hence, the sum of all terms, till $1000,$ will be zero.
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}=0$
View full question & answer→MCQ 761 Mark
If $x^2+p x+1=0$ and $(a-b) x^2+(b-c) x+(c-a)=0$ have both roots common, then what is the form of $a, b, c?$
- A
$a, b, c$ are in $A.P$
- ✓
$b, a, c$ are in $A.P$
- C
$b, a, c$ are in $G.P$
- D
$b, a, c$ are in $H.P$
AnswerCorrect option: B. $b, a, c$ are in $A.P$
Given, $(a-b) x^2+(b-c) x+(c-a)=0$ and $x^2+p x+1=0$
So, $\frac{1}{(\text{a}-\text{b})}=\frac{p}{(\text{b}-\text{c})}=\frac{1}{(\text{c}-\text{a})}$
Equating the above equation, we get,
$(b – c) = p(a – b)$ and
$(b – c) = p(c – a)$
So, $p(a – b) = p(c – a)$
$\Rightarrow a – b = c – a$
So, $2a = b + c$ which means that $b, a, c$ are in $A.P.$
View full question & answer→MCQ 771 Mark
A real value of $x$ satisfies the equation $\frac{3-4\text{ix}}{3+4\text{ix}}=\text{a}-\text{ib}\Big(\text{a},\text{b}\in\text{R}\Big),$ if $\text{a}^2+\text{b}^2=$
Answer$\text{a}-\text{ib}=\frac{3-4\text{ix}}{3+4\text{ix}}$
$=\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}$
$=\frac{9+16\text{x}^2\text{i}^2-24\text{xi}}{9-16\text{x}^2\text{i}^2}$
$=\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}$
$\Rightarrow\ |\text{a}-\text{ib}|^2=\Bigg|\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}\Bigg|^2$
$\Rightarrow\text{a}^2+\text{b}^2=\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{(9+16\text{x}^2)^2}{(9+16\text{x}^2)^2}$
$=1$
View full question & answer→MCQ 781 Mark
If $p$ and $q$ are the roots of the equation $x^2+p x+q=0$ then, what are the values of $p$ and $q?$
- ✓
$p = 1, q = -2$
- B
$p = 0, q = 1$
- C
$p = -2, q = 0$
- D
$p = -2, q = 1$
AnswerCorrect option: A. $p = 1, q = -2$
Since, $p$ and $q$ are the roots of the equation $x^2+p x+q=0$
So, $p + q = -p$ and $pq = q$
So, $pq = q$
And, $q = 0$ or $p = 1$
If, $q = 0$ then, $p = 0$ and if $p = 1$ then $q = -2.$
View full question & answer→MCQ 791 Mark
If, $x^4+4 x^3+6 a x^2+6 b x+c$ is divisible by $x^3+3 x^2+9 x+3$. Then, what is the value of $a + b + c ?$
AnswerHere, $f(x) =\text{x}^4 + 4\text{x}^3 + 6\text{ax}^2 + 6\text{bx}+\text{c}$
so, let its roots be, $\alpha,\beta,\gamma,\delta$ and
$g(x) =\text{x}^3 + 3\text{x}^2 +9\text{x} + 3$
so, let its roots be, $\alpha,\beta,\gamma$
So, from here we can conclude,
$\alpha+\beta+\gamma+\delta=-4$ and $\alpha+\beta+\gamma+\delta=-3$
Thus, $\delta=-1$
This means, $(\text{x} + 1)(\text{x}^3 + 3\text{x}^2 + 9\text{x} + 3)$
On solving this equation in simpler form we get,
$\text{x}^4 + 4\text{x}^3 + 12\text{x}^2 + 12\text{x} + 3$
$\Rightarrow6\text{a}=12$
$\Rightarrow\text{a}=2$
$\Rightarrow6\text{b}=12$
$\Rightarrow\text{b}=2$
$\Rightarrow\text{c}=3$
$\Rightarrow\text{a}+\text{b}+\text{c}=7$
View full question & answer→MCQ 801 Mark
Which is the largest negative integer which satisfies $\frac{(\text{x}^2 – 1)}{(\text{x} – 2)(\text{x} – 3)}$ ?
Answer$\frac{(\text{x}^2 – 1)}{(\text{x} – 2)(\text{x} – 3)} > 0$
So, $x = -1, 1, 2, 3$
Thus, $\text{x}\in(-\infty, -1)\cup(1,\text{ 2})\cup(3,\infty)$
Therefore, the largest negative integer is $-2.$
View full question & answer→MCQ 811 Mark
If $\alpha$ and $\beta$ are imaginary cube roots of unity, then the value of $\alpha^4+\beta^{28}+\frac{1}{\alpha\beta}$ is:
View full question & answer→MCQ 821 Mark
The number of real solutions of $\left|2 x-x^2-3\right|=1$ is:
Answer$i.$Given equation: $\left|2 x-x^2-3\right|=1$
$2 x-x^2-3=1$
$\Rightarrow 2 x-x^2-4=0$
$\Rightarrow x^2-2 x+4=0$
$\Rightarrow(x-2)^2=0$
$\Rightarrow x=2,2$
$ii. -2 x+x^2+3=1$
$\Rightarrow x^2-2 x+2=0$
$\Rightarrow x^2-2 x+1+1=0$
$\Rightarrow(x-1+i)(x-1-i)=0$
$\Rightarrow x=1-i, 1+i$
Hence, the real solutions are $2,2 .$
View full question & answer→MCQ 831 Mark
In $z = 4 + i,$ what is imaginary part?
AnswerIn $z = a + b i, a$ is real part and $b$ is imaginary part.
So, in $4 + i,$ imaginary part is $1.$
View full question & answer→MCQ 841 Mark
Roots of a quadratic equation are real when discriminant is $............$
- A
- B
- C
- ✓
greater than or equal to zero
AnswerCorrect option: D. greater than or equal to zero
For a quadratic equation, $a x^2+b x+c=0$, discriminant is $b^2-4 a c$.
Roots are $\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ For real roots,
radical must be non$-$negative
i.e. discriminant should be greater than or equal to zero.
View full question & answer→MCQ 851 Mark
If $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})},$ then $|\text{z}|=$
- A
$1$
- ✓
$\frac{1}{\sqrt{26}}$
- C
$\frac{5}{\sqrt{26}}$
- D
AnswerCorrect option: B. $\frac{1}{\sqrt{26}}$
Let $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}-3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}+3}$
$\Rightarrow\text{z}=\frac{1}{5+\text{i}}\times\frac{5-\text{i}}{5-\text{i}}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25-\text{i}^2}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25+1}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{26}$
$\Rightarrow\text{z}=\frac{5}{26}-\frac{\text{i}}{26}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{676}+\frac{1}{676}}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{26}}$
View full question & answer→MCQ 861 Mark
If the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$ have a non$-$zero common roots, then $\lambda=$
AnswerLet a be the common roots of the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$
Therefore
$\alpha^2+2\text{a}+3\lambda=0\ ...(1)$
$2\alpha^2+3\alpha+5\lambda=0\ ...(2)$
Solving $(1)$ and $(2)$ by cross multiplication, we get
$\frac{\alpha^2}{10\lambda-9\lambda}=\frac{\alpha}{6\lambda-5\lambda}=\frac{1}{3-4}$
$\Rightarrow\text{a}^2=-\lambda,\alpha=-\lambda$
$\Rightarrow-\lambda=\lambda^2$
$\Rightarrow\lambda=-1$
View full question & answer→MCQ 871 Mark
If $n$ is a positive integer, then $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)4\text{n}+1$ is equal to:
View full question & answer→MCQ 881 Mark
If $\theta$ is the amplitude of $\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}},$ then $\tan\theta=$
- A
$\frac{2\text{a}}{\text{a}^2+\text{b}^2}$
- ✓
$\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
- C
$\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
- D
AnswerCorrect option: B. $\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\text{z}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\times\frac{\text{a}+\text{ib}}{\text{a}+\text{ib}}$
$\Rightarrow\text{z}=\frac{\text{a}^2+\text{i}^2\text{b}^2+2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2+2\text{abi}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}+\text{i}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{Re(z)}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2},\text{Im(z)}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
Since, $z$ lies in the first quadrant .
Therefore,
$\text{arg(z)}=\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
$\tan\theta=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
View full question & answer→MCQ 891 Mark
If $\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6},$ then
- A
$|\text{z}|=1,\text{arg(z)}=\frac{\pi}{4}$
- B
$|\text{z}|=1,\text{arg(z)}=\frac{\pi}{6}$
- C
$|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\frac{5\pi}{24}$
- ✓
$|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
AnswerCorrect option: D. $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
$\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{2}}+\frac{1}{2}\text{i}$
$\Rightarrow|\text{z}|=\sqrt{\Big(\frac{1}{\sqrt{2}}\Big)^2+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{1}{2}+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{3}{4}}$
$\Rightarrow|\text{z}|=\frac{\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
Since, the point z lies in the first quadrant.
Therefore, $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 901 Mark
The value of $\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$ is:
AnswerCorrect option: A. $\frac{1}{2}(1+\text{i})$
$\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$
$=\frac{\text{i}-1-\text{i}+1+\text{i}}{1+\text{i}} \ [$ As, $\text{i}^5=\text{i},\text{i}^6=-1,\text{i}^7=-\text{i},\text{i}^8=1,\text{i}^9=\text{i}]$
$=\frac{\text{i}}{\text{i}+1}$
$=\frac{\text{i}}{\text{i}+1}\times\frac{\text{i}-1}{\text{i}-1}$
$=\frac{\text{i}(\text{i}-1)}{\text{i}^2-1}$
$=\frac{\text{i}^2-\text{i}}{-2}$
$=\frac{1}{2}(1+\text{i})$
View full question & answer→MCQ 911 Mark
If $\alpha,\beta$ are the roots of the equation $a x^2+b x+c=0$, then $\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}$
- A
$\frac{\text{c}}{\text{ab}}$
- B
$\frac{\text{a}}{\text{bc}}$
- ✓
$\frac{\text{b}}{\text{ac}}$
- D
AnswerCorrect option: C. $\frac{\text{b}}{\text{ac}}$
Given equation: $a x^2+b x+c=0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\frac{\text{b}}{\text{a}}$
Product of the roots $=\alpha\beta=\frac{\text{c}}{\text{a}}$
$\therefore\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}=\frac{\text{a}\beta+\text{b}+\text{a}\alpha+\text{b}}{(\text{a}\alpha+\text{b})(\text{a}\beta+\text{b})}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}\alpha+\text{ab}\beta+\text{b}^2}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}(\alpha+\beta)+\text{b}^2}$
$=\frac{\text{a}\big(-\frac{\text{b}}{\text{a}}\big)+2\text{b}}{\text{a}^2\big(\frac{\text{c}}{\text{a}}\big)+\text{ab}\big(-\frac{\text{b}}{\text{a}}\big)+\text{b}^2}$
$=\frac{\text{b}}{\text{ac}}$
View full question & answer→MCQ 921 Mark
The value of $\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$ is
Answer$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$
$=\frac{\text{i}^{4\times148}+\text{i}^{4\times147+2}+\text{i}^{4\times147}+\text{i}^{4\times146+2}+\text{i}^{4\times146}}{\text{i}^{4\times145+2}+\text{i}^{4\times145}+\text{i}^{4\times144+2}+\text{i}^{4\times144}+\text{i}^{4\times143+2}}-1$ $[\because\text{i}^4=1$ and $\text{i}^2=-1]$
$=\frac{1+\text{i}^2+1+\text{i}^2+1}{\text{i}^2+1+\text{i}^2+1+\text{i}^2}-1$
$=\frac{1}{-1}-1$
$=-2$
View full question & answer→MCQ 931 Mark
If $x^2+a x+b=0$ and $x^2+b x+a=0$ have exactly $1$ common root then what is the value of $(a + b)$
AnswerSubtracting the equation $x^2+a x+b=0$ to $x^2+b x+a=0$ by solving the equation simultaneously, we get,
$(a – b)x + (b – a) = 0$
So, $(a – b) x = (a – b)$
Therefore, $x = 1$
Now, putting the value of $x = 3$ in any one of the equation, we get,
$1 + a + b = 0$
Therefore, $a + b = -1.$
View full question & answer→MCQ 941 Mark
Solve $2\text{x}^2+\sqrt{2\text{x}+2}=0$
- ✓
$\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
- B
$\frac{1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
- C
$\frac{1\pm\sqrt{7}}{2\sqrt{2}}$
- D
$\frac{-1\pm\sqrt{7}}{2\sqrt{2}}$
AnswerCorrect option: A. $\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
$2\text{x}^2+\sqrt{2\text{x}+2}=0$
$\Rightarrow\text{D}=(\sqrt{2})^2-4.2.2=2-16=-14$
Since $\text{D}\leq0,$ imaginary roots are there.
$\Rightarrow\text{x}=\frac{-\sqrt{2\pm}\sqrt{\text{D}}}{2.2}$
$=\frac{-\sqrt{2\pm\text{i}}\sqrt{\text{D}}}{2.2}$
$=\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
View full question & answer→MCQ 951 Mark
If $\left|z_1\right|=4,\left|z_2\right|=3$, then what is the value of $\left|z_1+z_2+3+4 i\right|$
- A
Less than $2$
- B
Less than $5$
- C
Less than $7$
- ✓
Less than $12$
AnswerCorrect option: D. Less than $12$
As, we know $|\text{Z}_2+\text{Z}_2+.....+\text{Z}_\text{n}|\leq|\text{Z}_1|+|\text{Z}_2+......+|\text{Z}_\text{n}|$
So, $ |\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq|\text{Z}_1| + |\text{Z}_2| + |3 + 4\text{i}|$
Now, putting the given values in the equation, we get,
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq4+3+\sqrt{9+16}$
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq4+3+5$
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq12$
View full question & answer→MCQ 961 Mark
The least positive integer $n$ such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer, is:
AnswerLet $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)$
$\Rightarrow\text{z}=\frac{2\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1+1} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{2}$
$\Rightarrow\text{z}=\text{i}-\text{i}^2$
$\Rightarrow\text{z}=\text{i}+1$
Now, $\text{z}^\text{n}=(1+\text{i})^\text{n}$
For $\text{n}=2,$
$\text{z}^2=(1+\text{i})^2$
$=1+\text{i}^2+2\text{i}$
$=1-1+2\text{i}$
$=2\text{i} \ ...(1)$
Since this is not a positive integer,
For $\text{n}=4,$
$\text{z}^4=(1+\text{i})^4$
$=\big[(1+\text{i})^2\big]^2$
$=(2\text{i})^2 [$Using $(1)]$
$=(4\text{i})^2$
$=-4 \ ...(2)$
This is a negative integer.
For $\text{n}=8,$
$\text{z}^8=(1+\text{i})^8$
$=\big[(1+\text{i})^4\big]^2$
$=(-4)^2 [$Using $(2)]$
$=16$
This is a positive integer.
Thus, $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is positive for $\text{n}=8.$
Therefore, $8$ is the least positive integer such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer.
View full question & answer→MCQ 971 Mark
The amplitude of $\frac{1}{\text{i}}$ is equal to:
- A
$0$
- B
$\frac{\pi}{2}$
- ✓
$-\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: C. $-\frac{\pi}{2}$
Let $\text{z}=\frac{1}{\text{i}}$
$\text{z}=\frac{1}{\text{i}}\times\frac{\text{i}}{\text{i}}$
$\text{z}=\frac{\text{i}}{\text{i}^2}$
$\text{z}=-\text{i}$
Since, $z (0, -1)$ lies on the negative imaginary axis .
Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
View full question & answer→MCQ 981 Mark
The common roots of the equations $x^{12}-1=0$ and $x^4+x^2+1=0$ are:
AnswerCorrect option: C. $\pm\text{w,}\pm\text{w}^2$
View full question & answer→MCQ 991 Mark
If $\text{z}=\text{a}+\text{ib}$ lies in third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant if:
- A
$\text{a}>\text{b}>0$
- B
$\text{a}<\text{b}<0$
- ✓
$\text{b}<\text{a}<0$
- D
$\text{b}>\text{a}>0$
AnswerCorrect option: C. $\text{b}<\text{a}<0$
Since, $\text{z}=\text{a}+\text{ib}$ lies in third quadrant.
$\Rightarrow\text{a}<0$ and $\text{b}<0 \ ...(1)$
Now,
$\frac{\bar{\text{z}}}{\text{z}}=\frac{\overline{\text{a}+\text{ib}}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}-\text{ib}}$
$=\frac{\text{a}^2+\text{i}^2\text{b}^2-2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$=\frac{\text{a}^2-\text{b}^2-2\text{abi}}{\text{a}^2+\text{b}^2}$
Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant.
$\Rightarrow\text{a}^2-\text{b}^2<0$
$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b})<0$
$\Rightarrow\text{a}-\text{b}>0$ and $\text{a}+\text{b}<0$
$\Rightarrow\text{a}>\text{b} \ ...(2)$
From $(1)$ and $(2),$
$\text{b}<\text{a}<0$
View full question & answer→MCQ 1001 Mark
The equation of the smallest degree with real coefficients having $1 + i$ as one of the roots is:
- A
$x^2+x+1=0$
- ✓
$x^2-2 x+2=0$
- C
$x^2+2 x+2=0$
- D
$x^2+2 x-2=0$
AnswerCorrect option: B. $x^2-2 x+2=0$
We know that, imaginary roots of a quadratic equation occur in conjugate pair.
It is given that, $1 + i$ is one of the roots.
So, the other root will be $1−i1 - i.$
Thus, the quadratic equation having roots $1 + i$ and $1 - i$ is,
$x^2-(1+i+1-i) x+(1+i)(1-i)=0$
$\Rightarrow x^2-2 x+2=0$
View full question & answer→