MCQ 11 Mark
The polar form of $(\text{i}^{25})^3$ is :
- A
$\cos\frac{\pi}{2}+\text{i}\sin\frac{\pi}{2}$
- B
$\cos\pi+\text{i}\sin\pi$
- C
$\cos\pi-\text{i}\sin\pi$
- ✓
$\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
AnswerCorrect option: D. $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
$(\text{i}^{25})^3=(\text{i})^{75}$
$=(\text{i})^{4\times18+3}$
$=(\text{i})^3$
$=-\text{i} \ (\because\text{i}^4=1)$
Let $\text{z}=0-\text{i}$
Since, the point $(0,−1)$ lies on the negative direction of imaginary axis.
Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
Modulus, $\text{r}=|\text{z}|=|1|=1$
$\therefore$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$
$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$
$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
View full question & answer→MCQ 21 Mark
The argument of $\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$ is :
- A
$60^\circ$
- B
$120^\circ$
- C
$210^\circ$
- ✓
$240^\circ$
AnswerCorrect option: D. $240^\circ$
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$
Rationalising the denominator,
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$=\frac{1+3\text{i}^2-2\sqrt{3}\text{i}}{1-3\text{i}^2}$
$=\frac{-2-2\sqrt{3}\text{i}}{4} \ (\because\text{i}^2=-1)$
$=\frac{-1}{2}-\text{i}\frac{\sqrt{3}}{2}$
Then, $\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=60^\circ$
Since the points $\big(\frac{-1}{2},-\frac{-\sqrt{3}}{2}\big)$ lie in the third quadrant, the argument is given by :
$\theta=180^\circ+60^\circ$
$=240^\circ$
View full question & answer→MCQ 31 Mark
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ then $\frac{1+\text{a}}{1-\text{a}}=$
AnswerCorrect option: C. $\text{i}\cot\frac{\theta}{2}$
$\text{a}=\cos\theta+\text{i}\sin\theta\ ($given$)$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-\text{i}\sin^2\theta+2\text{i}\sin\theta-\cos^2\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-(\sin^2\theta+\cos^2\theta)+2\text{i}\sin\theta}{1+(\sin^2\theta+\cos^2\theta)-2\cos\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\frac{\theta}{2}-\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{\text{i}\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\text{i}\cot\frac{\theta}{2}$
View full question & answer→MCQ 41 Mark
The complex number $z$ which satisfies the condition $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$ lies on:
- A
Circle $x^2+ y^2 = 1$
- ✓
The $x-$axis
- C
The $y-$axis
- D
The line $x + y = 1$
AnswerCorrect option: B. The $x-$axis
$\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
$\Rightarrow \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|^2=1^2$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\overline{\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)}=1$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\Big(\frac{-\text{i}+\overline{\text{z}}}{-\text{i}-\overline{\text{z}}}\Big)=1$
$\Rightarrow\Big(\frac{-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}{-\text{i}^2+\text{z}\text{i}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}\Big)=1$
$\Rightarrow-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}=-\text{i}^2+\text{zi}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}$
$\Rightarrow-\text{zi}+\overline{\text{z}}\text{i}=\text{zi}-\overline{\text{z}}\text{i}$
$\Rightarrow\overline{\text{z}}\text{i}+\overline{\text{z}}\text{i}=\text{zi}-\text{zi}$
$\Rightarrow2\overline{\text{z}}\text{i}=2\text{zi}$
$\Rightarrow\overline{\text{z}}=\text{z}$
$ \Rightarrow\text{z}$ is purely real.
View full question & answer→MCQ 51 Mark
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. $|\text{z}|^2=|\text{z}|^2$
It is obvious that, for any complex number $z,$
$|\text{z}|^2=|\text{z}|^2$
View full question & answer→MCQ 61 Mark
If $\text{z}=1-\cos\theta+\text{i}\sin\theta,$ then $|\text{z}|=$
AnswerCorrect option: C. $2\Big|\sin\frac{\theta}{2}\Big|$
$\therefore\text{z}=1-\cos\theta+\text{i}\sin\theta$
$\Rightarrow|\text{z}|=\sqrt{(1-\cos\theta)^2+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+1-2\cos\theta}$
$\Rightarrow|\text{z}|=\sqrt{2(1-2\cos\theta)}$
$\Rightarrow|\text{z}|=\sqrt{4\sin^2\frac{\theta}{2}}$
$\Rightarrow|\text{z}|=2\Big|\sin\frac{\theta}{2}\Big|$
View full question & answer→MCQ 71 Mark
If $\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i}),$ then $\text{x}^2+\text{y}^2=$
Answer$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$
Taking modulus on both the sides:
$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$
$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=100$
View full question & answer→MCQ 81 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 91 Mark
$\big(\sqrt{-2}\big)\big(\sqrt{-3}\big)$ is equal to :
- A
$\sqrt{6}$
- ✓
$-\sqrt{6}$
- C
$\text{i}\sqrt{6}$
- D
AnswerCorrect option: B. $-\sqrt{6}$
$\sqrt{-2}\times\sqrt{-3}$
$=\sqrt{2}\times\sqrt{3}\times\sqrt{-1}\times\sqrt{-1}$
$=\sqrt{6}\times\text{i}\times\text{i}$
$=\sqrt{6}\times\text{i}^2$
$=-\sqrt{6} \ [\because\text{i}^2=-1]$
View full question & answer→MCQ 101 Mark
If $\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number and $0 < \theta < 2\pi,$ then $\theta=$
- ✓
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerGiven :
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number
On rationalising, we get,
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}\times\frac{{1+2\text{i}\sin\theta}}{{1+2\text{i}\sin\theta}}$
$=\frac{(3+2\text{i}\sin\theta)({1+2\text{i}\sin\theta})}{(1)^2-(2\text{i}\sin\theta)^2}$
$=\frac{3+2\text{i}\sin\theta+{6\text{i}\sin\theta}+4\text{i}^2\sin^2\theta}{1+4\sin^2\theta}$
$=\frac{3-4\text{i}\sin^2\theta+{8\text{i}\sin\theta}}{1+4\sin^2\theta} \ [\because\text{i}^2=-1]$
$=\frac{3-4\text{i}\sin^2\theta}{1+4\sin^2\theta}+\text{i}\frac{8\sin\theta}{1+4\sin^2\theta}$
For the above term to be real, the imaginary part has to be zero.
$\therefore\frac{8\sin\theta}{1+4\sin^2\theta}=0$
$\Rightarrow8\sin\theta=0$
For this to be zero,
$\sin\theta=0$
$\Rightarrow\theta=0,\pi,2\pi,3\pi...$
But $0<\theta<2\pi$
Hence, $\theta=\pi$
View full question & answer→MCQ 111 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 121 Mark
If $\text{z}=\frac{-2}{1+\sqrt{3}},$ then the value of 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 131 Mark
If $\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy},$ then possible value of $\sqrt{\text{a}-\text{ib}}$ is :
AnswerCorrect option: D. $\text{x}-\text{iy}$
$\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy}$
Squaring on both the sides, we get,
$\text{a}+\text{ib}=\text{x}^2+(\text{iy})^2+2\text{i xy}$
$\Rightarrow\text{a}+\text{ib}=(\text{x}^2-\text{y}^2)+2\text{i xy}$
$\therefore\text{a}=(\text{x}^2-\text{y}^2)$
and $\text{b}=2\text{ xy}$
$\therefore\text{a}-\text{ib}=(\text{x}^2-\text{y}^2)-2\text{i xy}$
$\Rightarrow\text{a}-\text{ib}=\text{x}^2+\text{i}^2\text{y}^2-2\text{i xy} \ [\because\text{i}^2=-1]$
Taking square root on both the sides, we get:
$\sqrt{\text{a}-\text{ib}}=\text{x}-\text{iy}$
View full question & answer→MCQ 141 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 151 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 161 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 171 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 181 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 191 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 201 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 211 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 221 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 231 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 241 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 251 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 261 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 271 Mark
If $\text{z}=\Big(\frac{1+2\text{i}}{1-(1-\text{i})^2}\Big),$ then $\text{arg(z)}$ equal :
- ✓
$0$
- B
$\frac{\pi}{2}$
- C
$\pi$
- D
Answer$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$
$\Rightarrow\text{z}=1$
Since point $(1,0)$ lies on the positive direction of real axis, we have :
$\text{arg(z) }= 0$
View full question & answer→MCQ 281 Mark
If $\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta},$ then $\text{Re(z)}=$
AnswerCorrect option: B. $\frac{1}{2}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+\cos^2\theta-2\cos\theta+\text{i}\sin^2\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+1-2\cos\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\text{Re(z)}=\frac{(1-\cos\theta)}{2(1-\cos\theta)}=\frac{1}{2}$
View full question & answer→MCQ 291 Mark
The amplitude of $\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$ is :
- A
$\frac{\pi}{3}$
- B
$-\frac{\pi}{3}$
- ✓
$\frac{\pi}{6}$
- D
$-\frac{\pi}{6}$
AnswerCorrect option: C. $\frac{\pi}{6}$
Let $\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}\times\frac{\sqrt{3}-\text{i}}{\sqrt{3}-\text{i}}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+2\text{i}-\sqrt{3}\text{i}^2}{3-\text{i}^2}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{2\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{\sqrt{3}}{2}+\frac{1}{2}\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{3}}$
$\Rightarrow\alpha=\frac{\pi}{6}$
Since, $z$ lies in the first quadrant.
Therefore, $\text{arg(z)}=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=\frac{\pi}{6}$
View full question & answer→MCQ 301 Mark
If $\text{a}=1+\text{i},$ then $a^2$ equals:
AnswerCorrect option: B. $2\text{i}$
$\text{a}=1+\text{i}$
On squaring both the sides, we get,
$\text{a}^2=(1+\text{i})^2$
$\Rightarrow\text{a}^2=1+\text{i}^2+2\text{i}$
$\Rightarrow\text{a}^2=1-1+2\text{i} \ (\because\text{i}^2=-1)$
$\Rightarrow\text{a}^2=2\text{i}$
View full question & answer→MCQ 311 Mark
The principal value of the amplitude of $(1 + i)$ is :
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{12}$
- C
$\frac{3\pi}{4}$
- D
$\pi$
AnswerCorrect option: A. $\frac{\pi}{4}$
Let $\text{z}=(1+\text{i})$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, $z$ lies in the first quadrant.
Therefore, $\text{arg(z)}=\frac{\pi}{4}$
View full question & answer→MCQ 321 Mark
If $(\text{x}+\text{iy})^{\frac{1}{3}}=\text{a}+\text{ib,}$ then $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=$
Answer$(\text{x}+\text{iy)}^{\frac{1}{3}}=\text{a}+\text{ib}$
Cubing on both the sides, we get :
$\text{x}+\text{iy}=(\text{a}+\text{ib})^3$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+(\text{ib})^3+3\text{a}^2\text{bi}+3\text{a}(\text{bi})^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+\text{i}^3\text{b}^3+3\text{a}^2\text{bi}+3\text{i}^2\text{ab}^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-\text{i}\text{b}^3+3\text{a}^2\text{bi}-3\text{ab}^2 \ (\because\text{i}^2=-1,\text{i}^3=-\text{i})$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-3\text{a}\text{b}^2+\text{i}(-\text{b}^3+3\text{a}^2\text{b})$
$\therefore\text{x}=\text{a}^3-3\text{a}\text{b}^2$ and $\text{y}=3\text{a}^2\text{b}-\text{b}^3$
or, $\frac{\text{x}}{\text{a}}=\text{a}^2-3\text{b}^2$ and $\frac{\text{y}}{\text{b}}=3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\text{a}^2-3\text{b}^2+3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=4\text{a}^2-4\text{b}^2$
View full question & answer→MCQ 331 Mark
If $\frac{1+7\text{i}}{(2-\text{i})^2},$ then :
AnswerCorrect option: D. $\text{amp(z)}=\frac{3\pi}{4}$
$\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{4-1-4\text{i}} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}$
$\Rightarrow\text{z}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{9-16\text{i}^2}$
$\Rightarrow\text{z}=\frac{3-28+25\text{i}}{9+16}$
$\Rightarrow\text{z}=\frac{-25+25\text{i}}{25}$
$\Rightarrow\text{z}=-1+\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, $z$ lies in the second quadrant.
Therefore, $\text{amp(z)}=\pi-\alpha$
$=\pi-\frac{\pi}{4}$
$=\frac{3\pi}{4}$
View full question & answer→MCQ 341 Mark
If $\text{z}=\frac{1}{(2+3\text{i})^2},$ then $|\text{z}|=$
- ✓
$\frac{1}{13}$
- B
$\frac{1}{5}$
- C
$\frac{1}{12}$
- D
AnswerCorrect option: A. $\frac{1}{13}$
Let $\text{z}=\frac{1}{(2+3\text{i})^2}$
$\Rightarrow\text{z}=\frac{1}{4+9\text{i}^2+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{4-9+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}\times\frac{-5-12\text{i}}{-5-12\text{i}}$
$\Rightarrow\text{z}=\frac{-5-12\text{i}}{25+144}$
$\Rightarrow\text{z}=\frac{-5}{169}-\frac{12\text{i}}{169}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{169^2}+\frac{144}{169^2}}$
$\Rightarrow|\text{z}|=\frac{1}{\sqrt{169}}$
$\Rightarrow|\text{z}|=\frac{1}{13}$
View full question & answer→MCQ 351 Mark
If $z$ is a non $-$ zero complex number, then $\Big|\frac{|\bar{\text{z}}}|^2{\text{z}}\bar{\text{z}}\Big|$ is equal to :
AnswerCorrect option: A. $\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
$\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big| \ \Big(\because\text{z}\bar{\text{z}}=|\text{z}|^2\Big)$
Let $\text{z}=\text{a}+\text{ib}$
$\Rightarrow|\text{z}|=\sqrt{\text{a}^2+\text{b}^2}$
Let $\bar{\text{z}}=\text{a}-\text{ib}$
$\Rightarrow|\bar{\text{z}}|=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big|$
$=\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
View full question & answer→MCQ 361 Mark
Which of the following is correct for any two complex numbers $z_1$ and $z_2?$
- ✓
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
- B
$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)\text{arg}(\text{z}_2)$
- C
$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
- D
$|\text{z}_1+\text{z}_2|\geq|\text{z}_1|+|\text{z}_2|$
AnswerCorrect option: A. $|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
since we know that
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)+\text{arg}(\text{z}_2)$ and
$|\text{z}_1+\text{z}_2|\le|\text{z}_1|+|\text{z}_2|$
View full question & answer→MCQ 371 Mark
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ then $\text{x}^2+\text{y}^2$ is equal to :
- ✓
$\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
- B
$\frac{(\text{a}+1)^2}{4\text{a}^2+1}$
- C
$\frac{(\text{a}^2-1)^2}{(4\text{a}^2-1)^2}$
- D
AnswerCorrect option: A. $\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
$\text{x}+\text{iy}=\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}$
Taking modulus on both the sides, we get :
$\sqrt{\text{x}^2+\text{y}^2}=\frac{(\text{a}^2+1)^2}{\sqrt{4\text{a}^2+1}}$
Squaring both sides, we get,
$\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{{4\text{a}^2+1}}$
View full question & answer→MCQ 381 Mark
The argument of $\frac{1-\text{i}}{1+\text{i}}$ is :
- ✓
$-\frac{\pi}{2}$
- B
$\frac{\pi}{2}$
- C
$\frac{3\pi}{2}$
- D
$\frac{5\pi}{2}$
AnswerCorrect option: A. $-\frac{\pi}{2}$
Let $\text{z}=\frac{1-\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1-\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2-2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1-2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{-2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
Since $,z$ lies on negative direction of imaginary axis .
Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
View full question & answer→MCQ 391 Mark
The value of $(1+\text{i})^4+(1-\text{i})^4$ is :
AnswerUsing $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$
$(1+\text{i})^4+(1-\text{i})^4$
$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$
$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$
$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$
$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$
$=8\text{i}^2$
$=-8$
View full question & answer→MCQ 401 Mark
If $\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}},$ then $\text{y}=$
- A
$\frac{9}{85}$
- B
$\frac{-9}{85}$
- ✓
$\frac{53}{85}$
- D
AnswerCorrect option: C. $\frac{53}{85}$
$\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}\times\frac{7+6\text{i}}{7+6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{21+53\text{i}+30\text{i}^2}{49-36\text{i}^2}$
$\Rightarrow\text{x}+\text{iy}=\frac{21-30+53\text{i}}{49+36}$
$\Rightarrow\text{x}+\text{iy}=\frac{-9}{85}+\text{i}\frac{53}{85}$
On comparing both the sides:
$\text{y}=\frac{53}{85}$
View full question & answer→MCQ 411 Mark
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals :
AnswerLet $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
View full question & answer→MCQ 421 Mark
If $\text{z}=\Big(\frac{1+\text{i}}{1-\text{i}}\Big),$ then $z^4$ equals:
AnswerLet $\text{z}=\frac{1+\text{i}}{1-\text{i}}$
Rationalising the denominator:
$\text{z}=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1+2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
$\Rightarrow\text{z}^4=\text{i}^4$
Since $\text{i}^2=-1,$ we have:
$\Rightarrow\text{z}^4=\text{i}^2\times\text{i}^2$
$\Rightarrow\text{z}^4=1$
View full question & answer→MCQ 431 Mark
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2\times5\times10\times...\times(1+\text{n}^2)$ is equal to :
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 written as $|(1+\text{i})\|(1+2\text{i})\|(1+3\text{i})|...|(1+\text{ni})|$
$\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{2}\times\sqrt{5}\times\sqrt{10}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get :
$2\times5\times10\times ... \times(1+\text{n}^2)=\text{a}^2+\text{b}^2$
View full question & answer→