2 Marks Questions

Question 12 Marks

If $z=x+i y$ and $\omega=\frac{1-i z}{z-i}$, then what does $|\omega|$ $=1$ show in a complex plane?

Answer

$\omega=\frac{1-i z}{z-i} \text { and }|\omega|=1$
$\text {then}\quad \left|\frac{1-i z}{z-i}\right|=1$
$\Rightarrow \quad|1-i z|=|z-i|$
$\Rightarrow \quad|1-i(x+i y)|=|x+i y-i|$
$\Rightarrow \quad|1+y-i x|=|x+i(y-1)|$
$\Rightarrow \quad \sqrt{(1+y)^2+(-x)^2}=\sqrt{x^2+(y-1)^2}$
$\Rightarrow \quad 1+y^2+2 y+x^2=x^2+y^2+1-2 y$
$\Rightarrow \quad 4 y=0$
$\Rightarrow \quad y=0$
Hence, $z=x+i y=x, z$ lies on the real axis.

View full question & answer→

Question 22 Marks

Answer

$\begin{array}{l}\text {Let}\quad z_1=x_1+i y_1 \\\quad \quad\quad z_2=x_2+i y_2\end{array}$
$\therefore \quad\left|z_1\right|=\sqrt{x_1^2+y_1^2}$ and $\left|z_2\right|=\sqrt{x_2^2+y_2^2}$
Given, $\left|z_1\right|=\left|z_2\right|$
$\therefore \sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$
$\Rightarrow \quad x_1^2+y_1^2=x_2^2+y_2^2$
$\Rightarrow \quad x_1^2=x_2^2 \quad$ and $\quad y_1^2=y_2^2$
$\Rightarrow \quad x_1= \pm x_2 \quad$ and $y_1= \pm y_2$
then, $\quad z_1=x_1+i y_1= \pm x_2 \pm i y_2$
$=x_2+i y_2,-x_2-i y_2,-x_2+i y_2, x_2-i y_2$
$\therefore \quad z_1=z_2$ and $z_1=-z_2$
So, it is not necessary that $z_1=z_2$.

View full question & answer→

Question 32 Marks

If $z$ is a complex number and $\bar{z}$ is its conjugate then prove that :
$z^{-1}=\frac{\bar{z}}{|z|^2} \text {, where } z \neq 0$

Answer

Let
$\begin{array}{c}z=x+i y \quad \Rightarrow \bar{z}=x-i y \\z^{-1}=\frac{1}{z}=\frac{1}{x+i y}=\frac{1}{x+i y} \times \frac{x-i y}{x-i y} \\=\frac{x-i y}{(x)^2-(i y)^2}=\frac{x-i y}{x^2+y^2} \\z^{-1}=\frac{\bar{z}}{|z|^2}\left[\because|z|=\sqrt{x^2+y^2}\right] \text { Hence proved. }\end{array}$

View full question & answer→

Question 42 Marks

Write the radius and centre of the circle $z \bar{z}-(2$ $+3 i) z-(2-3 i) \bar{z}+9=0$ where $z=x+ iy$.

Answer

$\begin{aligned}\text {Let}\quad z =x+i y \end{aligned}$
$\text {then}\quad\bar{z}=x-i y $
Given equation of circle
$z \bar{z}-(2+3 i) z-(2-3 i) \bar{z}+9=0$
$\begin{array}{r}\Rightarrow \quad(x+i y)(x-i y)-(2+3 i)(x+i y)-(2-3 i) (x-i y)+9=0\end{array}$
$\begin{array}{r}\Rightarrow \quad x^2-i^2 y^2-x(2+3 i)-i y(2+3 i)-x(2-3 i) +i y(2-3 i)+9=0\end{array}$
$\begin{array}{l} \Rightarrow \quad x^2+y^2+x(-2-3 i-2+3 i)+i y (2-3 i-2-3 i)+9=0\left[\because i^2=-1\right]\end{array}$
$\begin{array}{l}\Rightarrow \quad x^2+y^2+x(-4)+i y(-6 i)+9=0 \\ \Rightarrow \quad x^2+y^2-4 x-6 y i^2+9=0 \\ \Rightarrow \quad x^2+y^2-4 x+6 y+9=0\ldots\ldots (1)\end{array}$
Comparing the equation (1) with the general equation of circle.
$x^2+y^2+2 g x+2 f y+c=0$
Required centre $=(-g,-f)=(2,-3)$
And the required radius
$\begin{array}{l}=\sqrt{g^2+f^2-c}=\sqrt{4+9-9} \\ =\sqrt{4}=2\end{array}$

View full question & answer→

Question 52 Marks

Find the value of the following :
$(1+i)^8+(1-i)^8$

Answer

$\begin{aligned}(1+i)^8+(1-i)^8 & =\left[(1+i)^2\right]^4+\left[(1-i)^2\right]^4 \\ & =\left[1+2 i+i^2\right]^4+\left[1-2 i+i^2\right]^4 \\ & =[1+2 i-1]^4+[1-2 i-1]^4 \\ & =[2 i]^4+[-2 i]^4=16 i^4+16 i^4 \\ & =16 \times 1+16 \times 1 \quad \because i^4=1 \\ & =16+16=32\end{aligned}$

View full question & answer→

Question 62 Marks

Write the real values of $x$ and $y$ for the equation $(1+i) y^2+(6+i)=(2+i) x$.

Answer

$\begin{array}{l}(1+i) y^2+(6+i)=(2+i) x \\\Rightarrow\quad\left(y^2+6\right)+i\left(y^2+1\right)=2 x+i x\end{array}$
comparing real and imaginary parts-
$\quad \quad y^2+6=2 x\ldots\ldots (1)$
and $y^2+1=x\ldots\ldots (2)$
$\begin{array}{l}\text {Subtracting (1) and (2) }\quad\quad\quad x=5 \\\text {Putting the value of x in (1)}\quad y^2+6=10 \\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad y^2=10-6 \\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad y^2=4 \Rightarrow y= \pm 2 \end{array}$
So, $x=5$ and $y= \pm~ 2$

View full question & answer→

Question 72 Marks

Find the value of $x$ and $y$ for the equation
$\frac{x-1}{3+i}+\frac{y-1}{3-i}=i$

Answer

$\begin{array}{l}\frac{x-1}{3+i}+\frac{y-1}{3-i}=i \\\Rightarrow \quad \frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}=i \\\Rightarrow \quad \frac{3 x-3-x i+i+3 y-3+i y-i}{(3)^2-(i)^2}=i \\\Rightarrow \quad \frac{(3 x+3 y-6)+i(y-x)}{9-i^2}=i \\\Rightarrow \quad \frac{3 x+3 y-6}{10}+i \frac{y-x}{10}=0+i\end{array}$
Separating real and imaginary parts :
$\begin{array}{ll}\therefore & \frac{3 x+3 y-6}{10}=0 \text { and } \frac{y-x}{10}=1 \\\Rightarrow & x+y=2 \text { and } x-y=-10 \\\Rightarrow & x=-4 \text { and } y=6\end{array}$

View full question & answer→

Question 82 Marks

Write the multiplicative inverse of complex number $(-2+3 i)$.

Answer

Given, $z=-2+3 i$
$\begin{aligned}\frac{1}{z} & =z^{-1}=\frac{1}{-2+3 i} \\\therefore \quad z^{-1} & =\frac{1}{(-2+3 i)} \times \frac{(-2-3 i)}{(-2-3 i)}\end{aligned}$
[Multiplying numerator and denominator by ( $-2-3 i$ ]
$\begin{aligned}z^{-1} & =\frac{-2-3 i}{(-2)^2-(3 i)^2} \\& =\frac{-2-3 i}{4+9} \\z^{-1} & =\frac{-2}{13}-\frac{3}{13} i\end{aligned}$

View full question & answer→

Question 92 Marks

Find the Simplest value of the following :
(a) $i^9$
(b) $i^{51}$
(c) $i^{-63}$
(d) $i^{342}$

Answer

(a) $i^9=i^8 \cdot i=\left(i^2\right)^4 \cdot i=(-1)^4 \cdot i=i$.
(b) $i^{51}=i^{50} \cdot i=\left(i^2\right)^{25} \cdot i=(-1)^{25} \cdot i=-i$.
(c) $i^{-63}=\frac{1}{i^{63}} \times \frac{i}{i}=\frac{i}{i^{64}}=\frac{i}{\left(i^4\right)^{16}}=i\left[\because i^4=1\right]$
(d) $i^{342}=\left(i^4\right)^{85} \cdot i^2=(1)^{85} \cdot i^2=1 \cdot i^2=1 \cdot-1=-1$.

View full question & answer→

Question 102 Marks

Write the conjugate of the following complex numbers :
$2 i, 3 i-5,7+11 i, 12 i+9 .$

Answer

Conjugate of $2 i=-2 i$
Conjugate of $3 i-5=-3 i-5$
Conjugate of $7+11 i=7-11 i$
Conjugate of $12 i+9=-12 i+9$.

View full question & answer→

Question 112 Marks

If $z \in C$ then write the locus of $|z-(3-4 i)|=7$ in Argand plane.

Answer

$\begin{array}{l}|z-(3-4 i)| \quad=7 \\|x+i y-3+4 i|=7 \\|(x-3)+i(y+4)|=7 \\\sqrt{(x-3)^2+(y+4)^2}=7 \\\therefore \quad(x-3)^2+(y+4)^2=49\end{array}$

View full question & answer→

Question 122 Marks

Write the complex number $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$ in the form of $a+i b$.

Answer

$\begin{aligned} & \frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}=\frac{6+9 i-4 i-6 i^2}{2-i+4 i-2 i^2} \\ = & \frac{6+5 i+6}{2+3 i+2}=\frac{12+5 i}{4+3 i} \\ = & \frac{(12+5 i) \times(4-3 i)}{(4+3 i) \times(4-3 i)}=\frac{48-36 i+20 i-15 i^2}{16-9 i^2} \\ = & \frac{48-16 i+15}{16+9}=\frac{63-16 i}{25} \\ = & \frac{63}{25}-\frac{16}{25} i\end{aligned}$

View full question & answer→

Maths 2 Marks Questions

Test yourself on this topic