[Om Sharma]

Answer

49

[Om Sharma]

Solution

$\mathrm{z}=2-\mathrm{i}\left(2 \tan \frac{5 \pi}{8}\right)=\mathrm{x}+\mathrm{iy}$ (let)

\[ \begin{aligned} & |z|=r=\sqrt{x^{2}+y^{2}}=\sqrt{(2)^{2}+\left(2 \tan \frac{5 \pi}{8}\right)^{2}}=\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right|=2 \sec \frac{3 \pi}{8}\left(\sec \frac{5 \pi}{8}<0\right) \\ & \arg (z)=\theta=\tan ^{-1}\left(\frac{y}{x}\right) \text { as } x, y>0=\tan ^{-1}\left(\frac{-2 \tan \frac{5 \pi}{8}}{2}\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right)=\tan ^{-1}\left(\tan \frac{3 \pi}{8}\right)=\frac{3 \pi}{8} \end{aligned} \]

[Om Sharma]

Answer

D

[Om Sharma]

Solution

let $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy} \Rightarrow \overline{\mathrm{z}}^{2}=\mathrm{x}^{2}-\mathrm{y}^{2}-2 \mathrm{ixy} \Rightarrow(\overline{\mathrm{z}})^{2}+|\mathrm{z}|=0$

\[ \Rightarrow x^{2}-y^{2}-2 i x y+\sqrt{x^{2}+y^{2}}=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}-2 x y i=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}=0 \text { and } x y=0 \]

Case $\mathrm{I}: \mathrm{x}=0 \quad$ Case II : $\mathrm{y}=0$

\[ \begin{array}{ll} -\mathrm{y}^{2}+|\mathrm{y}|=0 \Rightarrow|\mathrm{y}|=|\mathrm{y}|^{2} & \mathrm{x}^{2}+|\mathrm{x}|=0 \\ \Rightarrow|\mathrm{y}|(|\mathrm{y}|-1)=0 & \Rightarrow \mathrm{x}=0 \\ \Rightarrow \mathrm{y}=0, \pm 1 & \Rightarrow \mathrm{z}=0 \end{array} \]

$\mathrm{z}=0, \mathrm{i},-\mathrm{i}$ are roots $\alpha=\mathrm{i}-\mathrm{i}=0 ; \beta=\mathrm{i}(-\mathrm{i})=1 \Rightarrow 4\left(\alpha^{2}+\beta^{2}\right)=4(0+1)=4$

[Om Sharma]

Answer

C

[Om Sharma]

Solution

$\mathrm{z}^{2}=-\mathrm{i} \overline{\mathrm{z}} \Rightarrow\left|\mathrm{z}^{2}\right|=|\mathrm{i} \overline{\mathrm{z}}|=|\mathrm{i} \| \overline{\mathrm{z}}|=|\mathrm{z}| \Rightarrow|\mathrm{z}|^{2}-|\mathrm{z}|=0 \Rightarrow|\mathrm{z}|(|\mathrm{z}|-1)=0$
$|z|=0$ (not acceptable as $\mathrm{xy} \neq 0$ ) $\therefore|\mathrm{z}|=1 \Rightarrow|\mathrm{z}|^{2}=1$

[Om Sharma]

Answer

D

[Om Sharma]

Solution

$\mathrm{x}^{2}-(3-2 \mathrm{i}) \mathrm{x}-(2 \mathrm{i}-2)=0$

\[ \begin{aligned} & x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^{2}-4(1)(-(2 i-2))}}{2(1)}=\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\ & x=\frac{3-2 i \pm \sqrt{(1)^{2}+(2 i)^{2}-2(1)(2 i)}}{2}=\frac{3-2 i \pm(1-2 i)}{2} \Rightarrow x=2-2 i \text { or } 1+0 i \end{aligned} \]

According to question $\alpha+\beta \mathrm{i}=2-2 \mathrm{i}$ and $\gamma+\delta \mathrm{i}=1+\mathrm{i} .0$ we get $\alpha=2, \beta=-2, \gamma=1, \delta=0$ So, $\alpha \gamma+\beta \delta=2(1)+(-2)(0)=2$

[Om Sharma]

Answer

D

[Om Sharma]

Solution

As $|\mathrm{z}|=\mathrm{r}=1$ so we have $\mathrm{z}=\mathrm{e}^{\mathrm{i} \theta}$ and $\frac{\mathrm{z}}{\overline{\mathrm{z}}}=\mathrm{e}^{\mathrm{i} 2 \theta}$

\[ \left|\frac{\mathrm{z}}{\overline{\mathrm{z}}}+\frac{\overline{\mathrm{z}}}{\mathrm{z}}\right|=\left|\mathrm{e}^{\mathrm{i} 2 \theta}+\mathrm{e}^{-\mathrm{i} 2 \theta}\right|=1 \Rightarrow|\cos 2 \theta|=\frac{1}{2} \]

From graph we can conclude that given equation has 8 solution if $\theta \in[0,2 \pi]^{\mathrm{l}}$

[Om Sharma]

Answer

A

[Om Sharma]

Solution

$\alpha+\beta=a, \alpha \beta=-b \Rightarrow P_{6}=a P_{5}+b P_{4} \Rightarrow 45 \sqrt{7} i=a \times 11 \sqrt{7} i+b(-3 \sqrt{7}) i$

\[ \begin{aligned} & 45=11 \mathrm{a}-3 \mathrm{~b} \quad \ldots(1) \text { and } \mathrm{P}_{5}=\mathrm{aP}_{4}+\mathrm{bP}_{3} \\ & 11 \sqrt{7} \mathrm{i}=\mathrm{a}(-3 \sqrt{7} \mathrm{i})+\mathrm{b}(-5 \sqrt{7} \mathrm{i}) \Rightarrow 11=-3 \mathrm{a}-5 \mathrm{~b} \quad \ldots(2) \Rightarrow \mathrm{a}=3, \mathrm{~b}=-4 \text { we get } \mathrm{x}^{2}-3 \mathrm{x}+4=0 \text { with roots } \alpha \text { and } \beta \\ & \alpha+\beta=3, \alpha \beta=4 \Rightarrow\left|\alpha^{4}+\beta^{4}\right|=\left|\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2}\right|=\left|\left((\alpha+\beta)^{2}-2 \alpha \beta\right)^{2}-2(\alpha \beta)^{2}\right|=\left|\left(3^{2}-8\right)^{2}-2 \cdot 4^{2}\right|=\|-32 \mid=31 \end{aligned} \]

[Om Sharma]

Answer

31

[Om Sharma]

Solution

$2 \mathrm{z}^{2}-3 \mathrm{z}-2 \mathrm{i}=0 \Rightarrow 2\left(\mathrm{z}-\frac{\mathrm{i}}{\mathrm{z}}\right)=3$

\[ \alpha-\frac{\mathrm{i}}{\alpha}=\frac{3}{2} \Rightarrow \alpha^{2}-\frac{1}{\alpha^{2}}-2 \mathrm{i}=\frac{9}{4} \Rightarrow \frac{9}{4}+2 \mathrm{i}=\alpha^{2}-\frac{1}{\alpha^{2}} \Rightarrow \frac{81}{16}-4+9 \mathrm{i}=\alpha^{4}+\frac{1}{\alpha^{4}}-2 \Rightarrow \frac{49}{16}+9 \mathrm{i}=\alpha^{4}+\frac{1}{\alpha^{4}} \]

Similarly $\frac{49}{16}+9 \mathrm{i}=\beta^{4}+\frac{1}{\beta^{4}} \Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right)+\beta^{15}\left(\beta^{4}+\frac{1}{\beta^{4}}\right)}{\alpha^{15}+\beta^{15}}=\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 \mathrm{i}\right)}{\left(\alpha^{15}+\beta^{15}\right)}=\frac{49}{16}+9 \mathrm{i}$ Real Part $=\frac{49}{16}$ and Imaginary Part $=9$ Required term $=16\left(\frac{49}{16}\right)(9)=49 \times 9=441$

[Om Sharma]

Answer

A

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