3 Marks Question

Question 13 Marks

If $x+i y=\frac{ C +i}{ C -i}$, where C is a real number, then prove that :

Answer

Given,
$\begin{aligned}x+i y & =\frac{C+i}{C-i} \\& =\frac{(C+i)(C+i)}{(C-i)(C+i)}\end{aligned}$
[Multiplying numerator and denominator by $( C +i)]$
$\begin{aligned}& =\frac{(C+i)^2}{(C)^2-(i)^2}=\frac{C^2+i^2+2 i C}{C^2+1} \\& =\frac{C^2-1+2 i C}{C^2+1} \\\Rightarrow \quad x+i y & =\left(\frac{C^2-1}{C^2+1}\right)+i\left(\frac{2 C}{C^2+1}\right)\ldots\ldots (1)\end{aligned}$
Compairing real and imaginary parts of equation (1)
$\begin{array}{l}x=\frac{C^2-1}{C^2+1}\ldots\ldots (2) \\y=\frac{2 C}{C^2+1}\ldots\ldots (3)\end{array}$
Adding the square of equation (2) and (3)
$\begin{aligned}x^2+y^2 & =\left(\frac{C^2-1}{C^2+1}\right)+\left(\frac{2 C}{C^2+1}\right)^2 \\& =\frac{\left(C^2-1\right)^2+4 C^2}{\left(C^2+1\right)^2}=\frac{\left(C^2+1\right)^2}{\left(C^2+1\right)^2}\end{aligned}$
$\therefore \quad x^2+y^2=1\quad\quad\quad\text {Hence proved.}$
Dividing equation (3) by equation (2)
$\begin{aligned}\quad \frac{y}{x} & =\frac{\frac{2 C}{C^2+1}}{\frac{C^2-1}{C^2+1}}=\frac{2 C}{C^2-1} \\\Rightarrow \quad \frac{y}{x} & =\frac{2 C}{C^2-1}\quad\quad\quad\text {Hence proved.}\end{aligned}$

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Question 23 Marks

Answer

$\begin{aligned} &\text {(i)}\quad z_1-\left.z_2\right|^2=\left(z_1-z_2\right)\left(\overline{z_1-z_2}\right) \quad\left(\because|z|^2=z \bar{z}\right) \\ & =\left(z_1-z_2\right)\left(\bar{z}_1-\bar{z}_2\right) \quad\left[\because \overline{z_1-z_2}=\bar{z}_1-\bar{z}_2\right] \\ & =z_1 \bar{z}_1-z_1 \bar{z}_2-z_2 \bar{z}_1+z_2 \bar{z}_2 \\ & =\left|z_1\right|^2-\left(z_1 \bar{z}_2+z_2 \bar{z}_1\right)+\left|z_2\right|^2 \\ & =\left|z_1\right|^2-\left(z_1 \bar{z}_2+\left(\overline{z_1 \bar{z}_2}\right)\right)+\left|z_2\right|^2 \quad \quad[\because \overline{(\bar{z})}=z] \\ & =\left|z_1\right|^2-2 \operatorname{Re}\left(z_1 \bar{z}_2\right)+\left|z_2\right|^2 \quad[\because z+\bar{z}=2 \operatorname{Re}(z)] \\ & \leq\left|z_1\right|^2+2\left|z_1 \bar{z}_2\right|+\left|z_2\right|^2 \quad[\because-2 \operatorname{Re}(z) \leq|z|] \\ & \Rightarrow\left|z_1-z_2\right|^2 \leq\left|z_1\right|^2+2\left|z_1\right|\left|\bar{z}_2\right|+\left|z_2\right|^2 \\ & \Rightarrow\left|z_1-z_2\right|^2 \leq\left|z_1\right|^2+2\left|z_1\right|\left|z_2\right|+\left|z_2\right|^2 \quad[\because|z|=|\bar{z}|] \\ & \Rightarrow\left|z_1-z_2\right|^2 \leq\left[\left|z_1\right|+\left|z_2\right|\right]^2 \\ & \therefore\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right| \quad \text { Hence proved. }\end{aligned}$
$\begin{aligned} &\text {(ii)}\quad\left|z_1+z_2\right|^2=\left(z_1+z_2\right)\left(\overline{z_1+z_2}\right) \quad\left(\because|z|^2=z \bar{z}\right) \\ & =\left(z_1+z_2\right)\left(\bar{z}_1+\bar{z}_2\right) \quad\left[\because \overline{z_1+z_2}=\bar{z}_1+\bar{z}_2\right] \\ & =z_1 \bar{z}_1+z_1 \bar{z}_2+z_2 \bar{z}_1+z_2 \bar{z}_2 \\ & =\left|z_1\right|^2+\left[z_1 \bar{z}_2+\left(\overline{z_1 \bar{z}_2}\right)\right]+\left|z_2\right|^2 \quad[\because \overline{(\bar{z})}=z] \\ & =\left|z_1\right|^2+2 \operatorname{Re}\left(z_1 \bar{z}_2\right)+\left|z_2\right|^2 \quad[\because z+\bar{z}=2 \operatorname{Re}(z)] \\ & \geq\left|z_1\right|^2-2\left|z_1 \bar{z}_2\right|+\left|z_2\right|^2 \\ & \Rightarrow\left|z_1+z_2\right|^2 \geq\left|z_1\right|^2-2\left|z_1\right|\left|\bar{z}_2\right|+\left|z_2\right|^2 \\ & \Rightarrow\left|z_1+z_2\right|^2 \geq\left|z_1\right|^2-2\left|z_1\right|\left|z_2\right|+\left|z_2\right|^2\end{aligned}$
$\begin{array}{l}\Rightarrow\left|z_1+z_2\right|^2 \geq\left[\left|z_1\right|-\left|z_2\right|\right]^2 \\ \therefore\left|z_1+z_2\right| \geq\left|z_1\right|-\left|z_2\right| \quad \text { Hence proved. }\end{array}$

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Question 33 Marks

If $|z|=1$, then prove that $\frac{z-1}{z+1},(z \neq-1)$ is a pure imaginary number. If $z=1$, then what conclusion do you draw from this?

Answer

Let $z=x+i y$
$\begin{array}{l}\therefore \quad|z|=\sqrt{x^2+y^2}=1 \Rightarrow x^2+y^2=1 \ldots\ldots (1)\\\begin{array}{l} \therefore \quad \frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y} \\\Rightarrow \quad \frac{(x-1)(x+1)+y^2+i y(x+1)-i y(x-1)}{(x+1)^2+y^2} \\=\frac{x^2-1+y^2+i y[x+1-x+1]}{x^2+y^2+2 x+1} \\\quad=\frac{x^2+y^2-1+2 i y}{x^2+y^2+1+2 x}\end{array}\end{array}$
Putting the value from equation (1)
$=\frac{1-1+2 i y}{1+1+2 x}=\frac{0+2 i y}{2(1+x)}$
$\begin{array}{l}=\frac{0}{2(1+x)}+\frac{2 i y}{2(1+x)} \\=0+\frac{y}{(1+x)} i\end{array}$
Hence, $\frac{z-1}{z+1}$ is a pure imaginary number when $|z|=1,(z \neq-1)$
Again If $z=1$ then $\frac{z-1}{z+1}=\frac{0}{1+1}=0$ which is a pure real number.

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