Download App

Maths 5 Marks Questions

Complex Numbers and Quadratic Equations · CBSE STD 11 Science (CBSE - English Medium). 12 questions.

Questions

5 Marks Questions

Take a timed test

12 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
If $|z_1| = 1(z_1 \neq -1)$ and $\text{z}_2=\frac{\text{z}_1-1}{\text{z}_1+1},$ then show that the real part of $z_2$ is zero.
Answer
Let $\text{z}=\text{x}+\text{yi}$
$|\text{z}_1|=\sqrt{\text{x}^2+\text{y}^2}=1$ [given that $|z_1| = 1]$
$\Rightarrow\text{x}^2+\text{y}^2=1$
Now$, \text{z}_2=\frac{\text{z}_1-1}{\text{z}_1+1}=\frac{\text{x}+\text{iy}-1}{\text{x}+\text{iy}+1}$
$=\frac{(\text{x}-1)+\text{iy}}{(\text{x}+1)+\text{iy}}=\frac{(\text{x}-1)+\text{iy}}{(\text{x}+1)+\text{iy}}\times\frac{\text{x}+1-\text{iy}}{\text{x}+1-\text{iy}}$
$=\frac{(\text{x}-1)(\text{x}+1)-\text{y}(\text{x}-1)\text{i}+\text{y}(\text{x}+1)\text{i}-\text{y}^2\text{i}^2}{(\text{x}+1)^2-\text{y}^2\text{i}^2}$
$=\frac{(1-1)}{\text{x}^2+\text{y}^2+2\text{x}+1}+\frac{2\text{y}}{\text{x}^2+\text{y}^2+2\text{x}+1}\text{i}$
$=0+\frac{2\text{y}}{\text{x}^2+\text{y}^2+2\text{x}+1}\text{i}$
Hence$,$ the real part of $z_2$ is $0$
View full question & answer→
Question 25 Marks
Find the complex number satisfying the equation $\text{z}+\sqrt{2}|(\text{z}+1)|+\text{i}=0$
Answer
We have $\text{z}+\sqrt{2}|(\text{z}+1)|+\text{i}=0\ ....(\text{i})$
Putting $\text{z}=\text{x}+\text{iy},$ we get
$\text{x}+\text{iy}+\sqrt{2}|\text{x}+\text{iy}+1|+\text{i}=0$
$\Rightarrow\text{x}+\text{i}(1+\text{y})+\sqrt{2}\Big[\sqrt{(\text{x}+1)^2+\text{y}^2}\Big]=0$
$\Rightarrow\text{x}+\text{i}(1+\text{y})+\sqrt{2}\sqrt{(\text{x}^2+2\text{x}+1+\text{y}^2)}=0$
Comparing real and imaginary parts to zero, we get
$\text{x}+\sqrt{2}\sqrt{\text{x}^2+2\text{x}+1+\text{y}^2}=0$
And $\text{y}+1=0$
$\Rightarrow\text{y}=-1$
Putting y = -1 into (ii), we get
$\text{x}+\sqrt{2}\sqrt{\text{x}^2+2\text{x}+1+1}=0$
$\Rightarrow\sqrt{2}\sqrt{\text{x}^2+2\text{x}+2}=-\text{x}$
$\Rightarrow2\text{x}^2+4\text{x}+4=\text{x}^2$
$\Rightarrow\text{x}^2+4\text{x}+4=0$
$\Rightarrow(\text{x}+2)^2=0$
$\Rightarrow\text{x}=-2$
$\therefore\ \text{z}=\text{x}+\text{iy}=-2-\text{i}$
View full question & answer→
Question 35 Marks
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ find the value of $\Big(\frac{1+\text{a}}{1-\text{a}}\Big)$
Answer
Given that, $\text{a}=\cos\theta+\text{i}\sin\theta$
$\therefore\ \frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$
$=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$=\frac{{1-\cos\theta+\text{i}\sin\theta}+\cos\theta-\cos^2\theta+\text{i}\sin\theta\cos\theta+\text{i}\sin\theta-\text{i}\sin\theta\cos\theta+\text{i}^2\sin^2\theta}{(1-\cos\theta)^2-\text{i}^2\sin^2\theta}$
$=\frac{1+\text{i}\sin\theta-\cos^2\theta+\text{i}\sin\theta-\sin^2\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$=\frac{2\text{i}\sin\theta}{2(1-\cos\theta)}=\frac{\text{i}\sin\theta}{1-\cos\theta}$
$=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\text{i}}{2\sin^2\frac{\theta}{2}}$
$=\cot\frac{\theta}{2}\text{i}$
Hence, $\frac{1+\text{a}}{1-\text{a}}=\text{i}\cot\frac{\theta}{2}$
View full question & answer→
Question 45 Marks
If $\frac{\text{z}-1}{\text{z}+1}$ is purely imaginary number (z ≠ -1), then find the value of |z|.
Answer
Given that $\frac{\text{z}-1}{\text{z}+1}$ is purely imaginary number.
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{x}+\text{iy}-1}{\text{x}+\text{iy}+1}\Big|=\frac{(\text{x}-1)+\text{iy}}{(\text{x}+1)+\text{iy}}$
$=\frac{(\text{x}-1)+\text{iy}}{(\text{x}+1)+\text{iy}}\times=\frac{(\text{x}+1)-\text{iy}}{(\text{x}+1)-\text{iy}}$
$\Rightarrow\frac{(\text{x}-1)(\text{x}+1)-\text{iy}(\text{x}-1)+(\text{x}+1)\text{iy}-\text{i}^2\text{y}^2}{(\text{x}+1)^2-\text{i}^2\text{y}^2}$
$\Rightarrow\frac{\text{x}^2-1+\text{iy}(\text{x}+1-\text{x}+1)+\text{y}^2}{\text{x}^2+1+2\text{x}+\text{y}^2}=\frac{\text{x}^2+\text{y}^2-1+2\text{yi}}{\text{x}^2+\text{y}^2+2\text{x}+1}$
$\Rightarrow\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+\text{y}^2+2\text{x}+1}+\frac{2\text{y}}{\text{x}^2+\text{y}^2+2\text{x}+1}$
Since, the number is purely imaginary, then real part = 0
$\therefore\ \frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+\text{y}^2+2\text{x}+1}=0$
$\Rightarrow\text{x}^2+\text{y}^2-1=0$
$\Rightarrow\text{x}^2+\text{y}^2=1$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=1$ $\therefore|\text{z}|=1$
View full question & answer→
Question 55 Marks
Write the complex number $\text{z}=\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}$ in polar form.
Answer
Given that,
$\text{z}=\frac{1-\text{i}}{\frac{1}{2}+\text{i}\frac{\sqrt{3}}{2}}=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}$
$=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{2-2\sqrt{3}\text{i}-2\text{i}+2\sqrt{3}\text{i}^2}{(1)^2-(\text{i}\sqrt{3})^2}=\frac{2-2\sqrt{3}\text{i}-2\text{i}-2\sqrt{3}}{1-3\text{i}^2}$
$=\frac{(2-2\sqrt{3})-(2+2\sqrt{3})\text{i}}{4}=\frac{1-\sqrt{3}}{2}-\frac{1+\sqrt{3}}{2}\text{i}$
$\Rightarrow\text{r}=\sqrt{\Big(\frac{1-\sqrt{3}}{2}\Big)^2+\Big(-\frac{1+\sqrt{3}}{2}\Big)^2}$
$=\sqrt{\frac{1+3-2\sqrt{3}}{4}+\frac{1+3+2\sqrt{3}}{4}}$
$=\sqrt{\frac{4-2\sqrt{3}+4+2\sqrt{3}}{4}}=\sqrt{\frac{8}{4}}$
So, $\text{r}=\sqrt{2}$
Now, $\text{arg(z)}=\tan^{-1}\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \theta=\tan^{-}\frac{\Big(\frac{1+\sqrt{3}}{2}\Big)}{\Big(\frac{1-\sqrt{3}}{2}\Big)}=\tan^{-1}\Big[-\Big(\frac{1+\sqrt{3}}{1-\sqrt{3}}\Big)\Big]$
$\Rightarrow\ \theta=\tan^{-1}\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\Rightarrow\ \theta=\tan^{-1}\Big[\tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)\Big]$ $\Bigg[\because\ \tan\Big(\frac{\pi}{4}+\frac{\pi}{6}\Big)=\frac{\tan\frac{\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{4}\tan\frac{\pi}{6}}\Bigg]$
$\Rightarrow\ \theta=\frac{5\pi}{12}$
Hence, the polar is
$\text{z}=\sqrt{2}\Big[\cos\Big(\frac{5\pi}{12}\Big)+\text{i}\sin\Big(\frac{5\pi}{12}\Big)\Big]$
View full question & answer→
Question 65 Marks
Show that the complex number z, satisfy the condition $\arg\Big(\frac{\text{z}-1}{\text{z}+1}\Big)=\frac{\pi}{4}$ lies on a circle.
Answer
Let $\text{z}=\text{x}+\text{iy}$
Given that, $\arg\Big(\frac{\text{z}-1}{\text{z}+1}\Big)=\frac{\pi}{4}$
$\Rightarrow\arg(\text{z}-1)-\arg(\text{z}+1)=\frac{\pi}{4}$ $\Big[\because\arg(\text{z}_1)-\arg(\text{z}_2)=\arg\frac{\text{z}_1}{\text{z}_2}\Big]$
$\Rightarrow\arg\big[\text{x}+\text{iy}-1\big]-\text{arg}\big[\text{x}+\text{iy}+1\big]=\frac{\pi}{4}$
$\Rightarrow\arg\big[(\text{x}-1)+\text{iy}\big]-\text{arg}\big[(\text{x}+1)+\text{iy}\big]=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\frac{\text{y}}{\text{x}-1}-\tan^{-1}\frac{\text{y}}{\text{x}+1}=\frac{\pi}{4}$ $\Big[\because\ \arg(\text{x}+\text{yi})=\tan^{-1}\frac{\text{y}}{\text{x}}\Big]$
$\Rightarrow\tan\Bigg(\frac{\frac{\text{y}}{\text{x}-\text{y}}-\frac{\text{y}}{\text{x}+1}}{1+\frac{\text{y}}{\text{x}-1}\times\frac{\text{y}}{\text{x}+1}}\Bigg)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{y}}{\text{x}^2+\text{y}^2-1}=1$
$\Rightarrow\text{x}^2+\text{y}^2-1=2\text{y}$
$\Rightarrow\text{x}^2+\text{y}^2-2\text{y}-1=0$ which is a circle.
Hence, z lies on a circle.
View full question & answer→
Question 75 Marks
If |z + 1| = z + 2(1 + i), then find z.
Answer
We have |z + 1| = z + 2(1 + i)
Putting z = x + iy, we get
Then, |x + iy + 1| = x + iy + 2(1 + i)
Putting |x + iy + 1| = x + iy + 2(1 + i)
Comparing real and imaginary parts, we get
$\sqrt{(\text{x}+1)^2+\text{y}^2}=\text{x}+2$
And $\text{y}+2=0$
$\Rightarrow\text{y}=-2$
Putting y = -2 into $\sqrt{(\text{x}+1)^2+\text{y}^2}=\text{x}+2,$ we get
$(\text{x}+1)^2+(-2)^2=(\text{x}+2)^2$
$\Rightarrow\text{x}^2+2\text{x}+1+4=\text{x}^2+4\text{x}+4$
$\Rightarrow2\text{x}=1$
$\therefore\ \text{z}=\text{x}+\text{iy}=\frac{1}{2}-2\text{i}$
View full question & answer→
Question 85 Marks
If arg(z - 1) = arg(z + 3i) then find x - 1, y where z = x + iy.
Answer
Given that, |z + 1| = arg(z + 3i)
⇒ arg[x + yi - 1] = arg[x + yi + 3i]
⇒ arg[(x - 1) + yi] = arg[x + (y + 3)i]
$\Rightarrow\tan^{-1}\frac{\text{y}}{\text{x}-1}=\tan^{-1}\frac{\text{y}+3}{\text{x}}$
$\Rightarrow\frac{\text{y}}{\text{x}-1}=\frac{\text{y}+3}{\text{x}}$
$\Rightarrow\text{xy}=(\text{x}-1)(\text{y}+3)$
$\Rightarrow\text{xy}=\text{xy}+3\text{x}-\text{y}-3$
$\Rightarrow3\text{x}-\text{y}=3$
$\Rightarrow3\text{x}-3=\text{y}$
$\Rightarrow3(\text{x}-1)=\text{y}$
$\Rightarrow\frac{(\text{x}-1)}{\text{y}}=\frac{1}{3}$
$\Rightarrow\text{x}-1:\text{ y}=1:3$
Hence, $\text{x}-1:\text{y}=1:3$
View full question & answer→
Question 95 Marks
If$ |z_1| = |z_2| = ........ = |z_n| = 1,$ then show that $|\text{z}_1+\text{z}_2+\text{z}_3+\ ....\ +\text{z}_\text{n}|=\Big|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}+\ .....\ +\frac{1}{\text{z}_\text{n}}\Big|$
Answer
We have $|\text{z}_1|=|\text{z}_2|=\ ....\ =|\text{z}_\text{n}|=1$
$\Rightarrow|\text{z}_1|^2=|\text{z}_2|^2=\ ....\ =|\text{z}_\text{n}|^2=1\ ....(\text{i})$
$\Rightarrow\text{z}_1\bar{\text{z}}_1=\text{z}_2\bar{\text{z}}_2=\ .....\ =\text{z}_\text{n}=\bar{\text{z}}_\text{n}=1$ $\Big[\because\text{z}\bar{\text{z}}=|\text{z}|^2\Big]$
$\Rightarrow\text{z}_1=\frac{1}{\bar{\text{z}}_1},\text{z}_2=\frac{1}{\bar{\text{z}}_2}=\ .....\ =\text{z}_\text{n}=\frac{1}{\bar{\text{z}}_\text{n}}$
$\text{L.H.S. }|\text{z}_1+\text{z}_2+\text{z}_3+\ ....\ +\text{z}_\text{n}|$
$=\Big|\frac{\text{z}_1\bar{\text{z}}_1}{\text{z}_1}+\frac{\text{z}_2\bar{\text{z}}_2}{\text{z}_2}+\frac{\text{z}_3\bar{\text{z}}_3}{\bar{\text{z}}_3}+\ .....\ +\frac{\text{z}_\text{n}\bar{\text{z}}_\text{n}}{\bar{\text{z}}_\text{n}}\Big|$
$=\bigg|\frac{|\text{z}_1|^2}{\bar{\text{z}}_1}+\frac{|\text{z}_2|^2}{\bar{\text{z}}_2}+\frac{|\text{z}_3|^2}{\bar{\text{z}}_3}+\ .....\ +\frac{|\text{z}_\text{n}|^2}{\bar{\text{z}}_\text{n}} \bigg|$ $\big[\text{z}\bar{\text{z}}=|\text{z}|^2\big]$
$=\Big|\frac{1}{\bar{\text{z}}_1}+\frac{1}{\bar{\text{z}}_2}+\frac{1}{\bar{\text{z}}_3}+\ .....\ +\frac{1}{\bar{\text{z}}_\text{n}}\Big| [$Using $(i)]$
$=\Big|\overline{\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}+\ ......\ +\frac{1}{\text{z}_\text{n}}}\Big|$ $\Big[\because\ \bar{\text{z}}_1+\bar{\text{z}}_2=\overline{\text{z}_1+\text{z}_2}\Big]$
$=\Big|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}+\ .....\ +\frac{1}{\text{z}_\text{n}}\Big|$ $\Big[\because|\text{z}|=|\bar{\text{z}}|\Big]$
$\text{L.H.S.}=\text{R.H.S.}$ Hence proved.
View full question & answer→
Question 105 Marks
Solve the equation$ |z| = z + 1 + 2i.$
Answer
Given that$, |z| = z + 1 + 2i$
$|z| = (z + 1) + 2i$
Squaring both sides
$|z|^2 = |z + 1|^2 + 4i^2 + 4(z + 1)i$
$\Rightarrow |z|^2 = |z|^2 + 1 + 2z - 4 + 4(z + 1)i$
$\Rightarrow 0 = -3 + 2z + 4(z + 1)i$
$\Rightarrow 3 - 2z - 4(z + 1)i = 0$
$\Rightarrow 3 - 2(x + yi) - 4[x + yi + 1]i = 0$
$\Rightarrow 3 - 2x - 2yi - 4xi - 4yi^2 - 4i = 0$
$\Rightarrow 3 - 2x + 4y - 2yi - 4i - 4xi = 0$
$\Rightarrow (3 - 2x + 4y) - i(2y + 4x + 4) = 0$
$\Rightarrow 3 - 2x + 4y = 0$
$\Rightarrow 2x - 4y = 3$
And $4x + 2y + 4 = 0$
$\Rightarrow 2x + y = -2$
Solving $eq. (i)$ and $(ii),$ we get
$\text{y}=-1$ and $\text{x}=-\frac{1}{2}$
Hence$,$ the value of $\text{z}=\text{x}+\text{yi}=\Big(-\frac{1}{2}-\text{i}\Big)$
View full question & answer→
Question 115 Marks
If for complex numbers $z_1$ and $z_2,$ arg $(z_1) -$ arg $(z_2) = 0,$ then show that $|z_1 - z_2| = |z_1| - |z_2|.$
Answer
Given that for$ z_1$ and $z_2,\  arg\  (z_1) - arg\  (z_2) = 0$
Let us represent $z_1$ and $z_2,$ in polar form.
$\text{z}_1=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)$ and $\text{z}_2=\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
$\text{arg}(\text{z}_1)=\theta_1$ and $\text{arg}(\text{z}_2)=0$
Since $\text{arg}(\text{z}_1)-\text{arg}(\text{z}_2)=0$
$\Rightarrow\theta_1-\theta_2=0$
$\Rightarrow\theta_1=\theta_2$
Now, $\text{z}_1-\text{z}_2=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)-\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
$=\text{r}_1\cos\theta_1+\text{i}\text{r}_1\sin\theta_1-\text{r}_2\cos\theta_1-\text{i}\text{r}_2\sin\theta_1$ $[\because\ \theta_1=\theta_2\big]$
$=(\text{r}_1\cos\theta_1-\text{r}_2\cos\theta_1)+\text{i}(\text{r}_1\sin\theta_1-\text{r}_2\sin\theta_1)$
$\therefore\ |\text{z}_1-\text{z}_2|=\sqrt{(\text{r}_1\cos\theta_1-\text{r}_2\cos\theta_1)^2+(\text{r}_1\sin\theta_1-\text{r}_2\sin\theta_1)^2}$
$=\sqrt{(\text{r}_1\cos^2\theta_1+\text{r}_2^2\cos^2\theta_1-2\text{r}_1\text{r}_2\cos^2\theta_1$
$+\text{r}_1^2\sin^2\theta_1+\text{r}^2_2\sin^2\theta_1+2\text{r}_1\text{r}_2\sin^2\theta_1)}$
$=\sqrt{\text{r}_1^2(\cos^2\theta_1+\sin^2\theta_1)+\text{r}^2_2(\cos^2\theta_1+\sin^2\theta_1)$
$-2\text{r}_1\text{r}_2(\cos^2\theta_1+\sin^2\theta_1)}$
$=\sqrt{\text{r}^2_1+\text{r}^2_2-2\text{r}_1\text{r}_2}=\sqrt{(\text{r}_1-\text{r}_2)^2}$
$=\text{r}_1-\text{r}_2$
$=|\text{z}_1|-|\text{z}_2|$
Hence, $|\text{z}_1-\text{z}_2|=|\text{z}_1|-|\text{z}_2|$
View full question & answer→
Question 125 Marks
Match the statements of Column $A$ and Column $B.$
  Column $A$   Column $B$
$a.$ The polar form of $\text{i}+\sqrt{3}$ is $i.$ Perpendicular bisector of segment joining $(– 2, 0)$ and $(2, 0).$
$b.$ The amplitude of $-1+\sqrt{-3}$ is $ii.$ On or outside the circle having centre at $(0, – 4)$ and radius $3.$
$c.$ If $|z + 2| = |z - 2|,$ then locus of $z$ is $iii.$ $\frac{2\pi}{3}$
$d.$ If $|z + 2i| = |z - 2i|,$ then locus of $z$ is $iv.$ Perpendicular bisector of segment joining $(0, – 2)$ and $(0, 2).$
$e.$ Region represented by $|\text{z}+4\text{i}|\geq3$ is $v.$ $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$
$f.$ Region represented by $|\text{z}+4|\leq3$ is $vi.$ On or inside the circle having centre $(– 4, 0)$ and radius $3$ units.
$g.$ Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in $vii.$ First quadrant.
$h.$ Reciprocal of $1 - i$ lies in $viii.$ Third quadrant.
Answer
  Column $A$   Column $B$
$a.$ The polar form of $\text{i}+\sqrt{3}$ is $v.$ $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$
$b.$ The amplitude of $-1+\sqrt{-3}$ is $ii.$ $\frac{2\pi}{3}$
$c.$ If $|z + 2| = |z - 2|,$ then locus of $z$ is $iii.$ Perpendicular bisector of segment joining $(-2, 0)$ and $(2, 0).$
$d.$ If $|z + 2i| = |z - 2i|,$ then locus of $z$ is $iv.$ Perpendicular bisector of segment joining $(0, -2)$ and $(0, 2).$
$e.$ Region represented by $|\text{z}+4\text{i}|\geq3$ is $v.$ On or outside the circle having centre at $(0, -4)$ and radius $3.$
$f.$ Region represented by $|\text{z}+4|\leq3$ is $vi.$ On or inside the circle having centre $(– 4, 0)$ and radius $3$ units.
$g.$ Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in $vii.$ Third quadrant.
$h.$ Reciprocal of $1 - i$ lies in $viii.$ First quadrant.
Solution:
  1. Given, $\text{z}=\text{i}+\sqrt{3}$
Polar form of $\text{z}=\text{r}\big[\cos\theta+\text{I}\sin\theta\big]$
$\Rightarrow\sqrt{3}+\text{i}=\cos\theta+\text{ri}\sin\theta$
$\Rightarrow\text{r}=\sqrt{(\sqrt{3})^2+(1)^2}=2$ And $\tan\alpha=\frac{1}{\sqrt{3}}$
$\Rightarrow\alpha=\frac{\pi}{6}$ Since $\text{x}>0,\text{ y}>0$
$\therefore$ Polar for of $\text{z}=2\Big[\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{6}\Big]$
  1. Given that $\text{z}=-1+\sqrt{3}=-1+\sqrt{3}\text{i}$
Here argument $(\text{z})=\tan^{-1}\Big|\frac{\sqrt{3}}{-1}\Big|=\tan^{-1}\big|\sqrt{3}\big|=\frac{\pi}{3}$
So, $\alpha=\frac{\pi}{3}$ Since $\text{x}<0$ and $\text{y}>0$
Then $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
  1. Given that, $|\text{z}+2|=|\text{z}-2|$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ |\text{x}+\text{yi}+2|=|\text{x}+\text{yi}-2|$
$\Rightarrow|(\text{x}+2)+\text{yi}|=|(\text{x}-2)+\text{yi}|$
$\Rightarrow\sqrt{(\text{x}+2)^2+\text{y}^2}=\sqrt{(\text{x}-2)^2+\text{y}^2}$
$\Rightarrow(\text{x}+2)^2+\text{y}^2=(\text{x}-2)^2+\text{y}^2$
$\Rightarrow(\text{x}+2)^2=(\text{x}-2)^2$
$\Rightarrow\text{x}^2+4+4\text{x}=\text{x}^2+4-4\text{x}$
$\Rightarrow8\text{x}=0$
$\Rightarrow\text{x}=0$
Which represent equation of $y-$axis and it is perpendicular to the line joining the points $(-2, 0)$ and $(2, 0).$
  1. $|\text{z}+2\text{i}|=|\text{z}-2\text{i}|$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ (\text{x}+\text{yi}+2\text{i})=|\text{x}+\text{yi}-2\text{i}|$
$\Rightarrow|\text{x}+(\text{y}+2)\text{i}|=|\text{x}+(\text{y}-2)\text{i}|$
$\Rightarrow\sqrt{\text{x}^2+(\text{y}+2)^2}=\sqrt{\text{x}^2+(\text{y}-2)^2}$
$\Rightarrow\text{x}^2+(\text{y}+2)^2=\text{x}^2+(\text{y}-2)^2$
$\Rightarrow(\text{y}+2)^2=(\text{y}-2)^2$
$\Rightarrow\text{y}^2+4+4\text{y}=\text{y}^2+4-4\text{y}$
$\Rightarrow8\text{y}=0$
$\Rightarrow\text{y}=0$
Which is the equation of $x-$axis and it is perpendicular to the line segment joining $(0, -2)$ and $(0, 2)$
  1. Given that, $|\text{z}+4\text{i}|\geq3$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ |\text{x}+\text{yi}+4|\geq3$
$\Rightarrow|\text{x}+(\text{y}+4)\text{i}|\geq3$
$\Rightarrow\text{x}^2+(\text{y}+4)^2\geq9$
$\Rightarrow\text{x}^2+\text{y}^2+8\text{y}+16\geq9$
$\Rightarrow\text{x}^2+\text{y}^2+8\text{y}+7\geq0$
$\Rightarrow\text{r}=\sqrt{(4)^2-7}=3$
Which represents a circle on or outside having centre $(0, -4)$ and radius $3.$
  1. $|\text{z}+4|\leq3$
Let $\text{z}=\text{x}+\text{yi}$ Then $|\text{x}+\text{yi}+4|\leq3$
$\Rightarrow|(\text{x}+4)+\text{yi}|\leq3$
$\Rightarrow\sqrt{(\text{x}+4)^2+\text{y}^2}\leq3$
$\Rightarrow\text{x}^2+8\text{x}+16+\text{y}^2\leq9$
$\Rightarrow\text{x}^2+\text{y}^2+8\text{x}+7\leq0$
Which is a circle having centre $(-4, 0)$ and $\text{r}=\sqrt{(4)^2-7}=\sqrt{9}=3$ and is on inside the circle.
  1. Let $\text{z}=\frac{1+2\text{i}}{1-\text{i}}$
$=\frac{1+2\text{i}}{1+\text{i}}\times\frac{1+\text{i}}{1+\text{i}}=\frac{1+\text{i}+2\text{i}+2\text{i}^2}{1-\text{i}^2}$
$=\frac{1+\text{i}+2\text{i}-2}{1+1}=\frac{-1+3\text{i}}{2}$
$=-\frac{1}{2}+\frac{3}{2}\text{i}$
$\therefore\ \bar{\text{z}}=-\frac{1}{2}-\frac{3}{2}\text{i}$
which lies in third quadrant.
  1. Given that, $\text{z}=1-\text{i}$
Reciprocal of $\text{z}=\frac{1}{\text{z}}=\frac{1}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}=\frac{1+\text{i}}{1+\text{i}^2}$
Which lies in first quadrant.
View full question & answer→
Generate Paper FreeAll Complex Numbers and Quadratic Equations topics