[Om Sharma]

all good

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Answer

A

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Solution

Let $\mathrm{z}_{1}=\left(\frac{\mathrm{z}-\overline{\mathrm{z}}+\mathrm{z} \overline{\mathrm{z}}}{2-3 \mathrm{z}+5 \overline{\mathrm{z}}}\right)$ Let $\mathrm{z}=3+\mathrm{i} y, \overline{\mathrm{z}}=3-\mathrm{iy} \Rightarrow \mathrm{z}_{1}=\frac{2 \mathrm{iy}+\left(9+\mathrm{y}^{2}\right)}{2-3(3+\mathrm{iy})+5(3-\mathrm{iy})}=\frac{9+\mathrm{y}^{2}+\mathrm{i}(2 \mathrm{y})}{8-8 \mathrm{iy}}=\frac{\left(9+\mathrm{y}^{2}\right)+\mathrm{i}(2 \mathrm{y})}{8(\mathrm{i}-\mathrm{iy})}$

\[ \begin{aligned} & \operatorname{Re}\left(z_{1}\right)=\frac{\left(9+y^{2}\right)-2 y^{2}}{8\left(i+y^{2}\right)}=\frac{9-y^{2}}{8\left(i+y^{2}\right)}=\frac{1}{8}\left[\frac{10}{i+y^{2}}-1\right] \because 1+y^{2} \in[1, \infty] \Rightarrow \frac{1}{i+y^{2}} \in(0,1] \Rightarrow \frac{10}{i+y^{2}}-1 \in(-1,9] \\ & \operatorname{Re}\left(z_{1}\right) \in\left(-\frac{1}{8}, \frac{9}{8}\right] \Rightarrow \alpha=-\frac{1}{8}, \beta=\frac{9}{8} \Rightarrow 24(\alpha-\beta)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30 \end{aligned} \]

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Answer

C

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Solution

$\mathrm{Z}=\frac{1+2 \mathrm{i} \sin \theta}{1-\mathrm{i} \sin \theta}$ is purely imaginary So, $\operatorname{Re}(\mathrm{z})=0 \Rightarrow \frac{1-2 \sin ^{2} \theta}{1+\sin ^{2} \theta}=0$
$\Rightarrow \sin ^{2} \theta=\frac{1}{2} \therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$, for $\theta \in(0,2 \pi) \therefore$ Sum of all values $=\frac{16 \pi}{4}=4 \pi$

[Om Sharma]

Answer

C

[Om Sharma]

Solution

$\mathrm{z}=\alpha+\mathrm{i} \beta \Rightarrow|\mathrm{z}+2|=\mathrm{z}+4(\mathrm{l}+\mathrm{i}) \Rightarrow \sqrt{(\alpha+2)^{2}+\beta^{2}}=(\alpha+4)+\mathrm{i}(\beta+4)$

\[ \begin{aligned} & \sqrt{(\alpha+2)^{2}+\beta^{2}}=\alpha+4 \quad \ldots(\mathrm{i}), \beta+4=0 \Rightarrow \beta=-4 \\ & \text { (i) } \Rightarrow \sqrt{(\alpha+2)^{2}+16}=\alpha+4 \Rightarrow \alpha^{2}+4 \alpha+20=\alpha^{2}+8 \alpha+16 \Rightarrow \alpha=1 \end{aligned} \]

$\alpha+\beta=-3, \alpha \beta=-4$ Equation with roots -3 and -4 is $\mathrm{x}^{2}+7 \mathrm{x}+12=0$

[Om Sharma]

Answer

A

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Solution

$\because \mathrm{z}=\mathrm{x}+\mathrm{iy} \mathrm{P}=\frac{2 \mathrm{z}-3 \mathrm{i}}{4 \mathrm{z}+2 \mathrm{i}} \Rightarrow \mathrm{P}=\frac{2 \mathrm{x}+(2 \mathrm{y}-3) \mathrm{i}}{4 \mathrm{x}+(4 \mathrm{y}+2) \mathrm{i}}$ is real number
$\mathrm{P}=\frac{(2 \mathrm{x}+(2 \mathrm{y}-3) \mathrm{i})(4 \mathrm{x}-(4 \mathrm{y}+2) \mathrm{i})}{(4 \mathrm{x})^{2}+(4 \mathrm{y}+2)^{2}}$ is real number, $\operatorname{lm}(\mathrm{P})=0 \Rightarrow-2 \mathrm{x}(4 \mathrm{y}+2)+(2 \mathrm{y}-3) 4 \mathrm{x}=0 \therefore \mathrm{x}=0$
Here $\mathrm{x}=0$ and $\mathrm{y} \in \mathrm{R}$ but $\mathrm{y} \neq-\frac{1}{2}$ is not acceptable in this case
Denominator will be zero. Hence, option (B) is correct.

[Om Sharma]

Answer

B

[Om Sharma]

Solution

$\mathrm{x}^{2}-\sqrt{2} \mathrm{x}+2=0 \Rightarrow \mathrm{x}=\frac{\sqrt{2} \pm \sqrt{6} \mathrm{i}}{2} \Rightarrow \alpha=\frac{\sqrt{2}+\sqrt{6} \mathrm{i}}{2}=\sqrt{2} \mathrm{e}^{\frac{\mathrm{i} \pi}{3}}$

\[ \begin{aligned} & \beta=\sqrt{2} \mathrm{e}^{-\frac{\mathrm{i} \pi}{3}} \Rightarrow \alpha^{14}=2^{7} \mathrm{e}^{\frac{\mathrm{i} 14 \pi}{3}}=128\left[\mathrm{e}^{\frac{\mathrm{i} 2 \pi}{3}}\right] \Rightarrow \beta^{14}=128\left[\mathrm{e}^{-\frac{\mathrm{i} 2 \pi}{3}}\right] \\ & \alpha^{14}+\beta^{14}=128(2) \cos \left(\frac{2 \pi}{3}\right)=-128 \end{aligned} \]

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Solution

Let $\mathrm{Z}=\mathrm{x}+\mathrm{iy}, \mathrm{x} \in \mathrm{R}, \mathrm{y} \in \mathrm{R} \Rightarrow \mathrm{x}-\mathrm{iy}=\mathrm{i}\left(\mathrm{x}^{2}-\mathrm{y}^{2}+(2 \mathrm{xy}) \mathrm{i}+\mathrm{x}\right)$
$\mathrm{x}=-2 \mathrm{xy} \ldots(1) \Rightarrow-\mathrm{y}=-\mathrm{y}^{2}+\mathrm{x}^{2}+\mathrm{x} \quad \ldots(2) \Rightarrow \mathrm{x}=0, \mathrm{y}=-\frac{1}{2}$ (from (1))
If $\mathrm{x} \neq 0$, then $\mathrm{y}=0,1 \Rightarrow$ If $\mathrm{y}=-\frac{1}{2}$, then $\mathrm{x}=\frac{1}{2},-\frac{3}{2} \Rightarrow \mathrm{Z}=0+\mathrm{i} 0,0+\mathrm{i}, \frac{1}{2}-\frac{\mathrm{i}}{2},-\frac{3}{2}-\frac{\mathrm{i}}{2}$

[Om Sharma]

Answer

D

[Om Sharma]

Answer

B

[Om Sharma]

Solution

$\operatorname{Re}\left(\mathrm{a} \mathrm{z}^{2}+\mathrm{bz}\right)=\mathrm{a} \Rightarrow \mathrm{az}^{2}+\mathrm{bz}+\mathrm{a} \overline{\mathrm{z}}^{2}+\mathrm{b} \overline{\mathrm{z}}=2 \mathrm{a} \Rightarrow \mathrm{a}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{b}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{a}$

\[ \operatorname{Re}\left(\mathrm{b} \mathrm{z}^{2}+\mathrm{az}\right)=\mathrm{b} \Rightarrow \mathrm{~b} \mathrm{z}^{2}+\mathrm{az}+\mathrm{b} \overline{\mathrm{z}}^{2}+\mathrm{a} \overline{\mathrm{z}}=2 \mathrm{~b} \Rightarrow \mathrm{~b}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{a}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{~b} \]

\[ \begin{aligned} & (1) \times b-(2) \times(a) \Rightarrow\left(b^{2}-a^{2}\right)(z+\bar{z})=0 \Rightarrow(z+\bar{z})=0\left(a^{2} \neq b^{2}\right) \\ & (1) \times a-(2) \times(b) \Rightarrow\left(a^{2}-b^{2}\right)(z+\bar{z})=2\left(a^{2}-b^{2}\right)\left(a^{2} \neq b^{2}\right) \\ & z^{2}+\bar{z}^{2}=2 \Rightarrow(z+\bar{z})^{2}-2 z \bar{z}=2 \Rightarrow z \bar{z}=-1 \Rightarrow 1+1^{2}=-1 \Rightarrow \text { No solution But when } a=-b, \operatorname{Re}\left(a z^{2}-a z\right)=a \\ & \Rightarrow \operatorname{Re}\left(a\left(x^{2}-y^{2}+i 2 x y\right)-a(x+i y)\right)=a \Rightarrow a\left(x^{2}-y^{2}\right)-a x=a \Rightarrow x^{2}-y^{2}-x=1 \Rightarrow x^{2}-x-1=y^{2} \end{aligned} \]

For any real values of y there two values of x, hence infinite complex numbers are possible.

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