[UPLIFTJEE]

Solution

For circle : $\left|\mathrm{z}-\mathrm{z}_{1}\right|^{2}+\left|\mathrm{z}-\mathrm{z}_{2}\right|^{2}=\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|^{2}$

\[ \begin{aligned} & r=\frac{\left|z_{1}-z_{2}\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \Rightarrow 2 \lambda=|\alpha-\beta|^{2} \Rightarrow|\alpha-\beta|=2 \sqrt{\lambda-1} \\ & |\alpha-\beta|^{2}=4 \lambda-4=2 \lambda \Rightarrow \lambda=2 \Rightarrow|\alpha-\beta|^{2}=4 \Rightarrow|\alpha-\beta|=2 \end{aligned} \]

[UPLIFTJEE]

Answer

2

[UPLIFTJEE]

Solution

According to the question,

\[ \begin{aligned} & \left|z_{1}-z_{0}\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}} \Rightarrow\left|z_{1}-z_{0}\right|=\sqrt{2} ; \text { centre }\left(\frac{1}{2}, \frac{3}{2}\right) \\ & z_{0}\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } z_{1}(1,1) \Rightarrow \tan \theta=-1 \Rightarrow \theta=135^{\circ} \\ & z_{2}\left(\frac{1}{2}+\sqrt{2} \cos 135^{\circ}, \frac{3}{2}+\sqrt{2} \sin 135^{\circ}\right) \text { or }\left(\frac{1}{2}-\sqrt{2} \cos 135^{\circ}, \frac{3}{2}-\sqrt{2} \sin 135^{\circ}\right) \\ & \Rightarrow z_{2}\left(-\frac{1}{2}, \frac{5}{2}\right) \text { or } z_{2}\left(\frac{3}{2}, \frac{1}{2}\right) \Rightarrow\left|z_{2}\right|^{2}=\frac{26}{4}, \frac{5}{2} \Rightarrow\left|z_{2}\right|_{\min }^{2}=\frac{5}{2} \end{aligned} \]

[UPLIFTJEE]

Answer

B

[UPLIFTJEE]

Solution

$\mathrm{w}=\mathrm{z} \overline{\mathrm{z}}+\mathrm{k}_{1} \mathrm{z}+\mathrm{k}_{2} \mathrm{iz}+\lambda(1+\mathrm{i}) \Rightarrow \operatorname{Re}(\mathrm{w})=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{k}_{1} \mathrm{x}-\mathrm{k}_{2} \mathrm{y}+\lambda=0$

\[ \begin{aligned} & \text { Centre } \equiv\left(-\frac{\mathrm{k}_{1}}{2}, \frac{\mathrm{k}_{2}}{2}\right) \equiv(1,2) \Rightarrow \mathrm{k}_{1}=-2, \mathrm{k}_{2}=4 \text { radius }=1 \Rightarrow \lambda=4 \\ & \operatorname{Im}(\mathrm{w})=\mathrm{k}_{1} \mathrm{y}+\mathrm{k}_{2} \mathrm{x}+\lambda=0 \therefore \text { line } \mathrm{AB}, 2 \mathrm{x}-\mathrm{y}+2=0 \\ & \mathrm{~d}=\left|\frac{2-2+2}{\sqrt{4+1}}\right|=\frac{2}{\sqrt{5}} \text { let } \mathrm{AB}=\ell, \mathrm{AM}^{2}=\mathrm{AC}^{2}-\mathrm{d}^{2} \\ & \frac{\ell^{2}}{4}=1-\frac{4}{5}=\frac{1}{5} \therefore 30 \ell^{2}=24 \end{aligned} \]

[UPLIFTJEE]

Answer

24

[UPLIFTJEE]

Solution

$\alpha=8-14 \mathrm{i}$, let $\mathrm{z}=\mathrm{x}+\mathrm{iy} \Rightarrow \mathrm{az}=(8 \mathrm{x}+14 \mathrm{y})+\mathrm{i}(-14 \mathrm{x}+8 \mathrm{y})$
$\mathrm{z}+\overline{\mathrm{z}}=2 \mathrm{x}, \mathrm{z}-\overline{\mathrm{z}}=2 \mathrm{iy}$ Set $\mathrm{A}: \frac{2 \mathrm{i}(-14 \mathrm{x}+8 \mathrm{y})}{\mathrm{i}(4 \mathrm{xy}-112)}=1 \Rightarrow(\mathrm{x}-4)(\mathrm{y}+7)=0 \Rightarrow \mathrm{x}=4$ or $\mathrm{y}=-7$
Set $\mathrm{B}: \mathrm{x}^{2}+(\mathrm{y}+3)^{2}=16$. When $\mathrm{x}=4, \mathrm{y}=-3$ when $\mathrm{y}=-7, \mathrm{x}=0 \therefore \mathrm{~A} \cap \mathrm{~B}=\{4-3 \mathrm{i}, 0-7 \mathrm{i}\}$
So, $\sum_{\mathrm{z} \in \mathrm{A} \cap \mathrm{B}}(\operatorname{Rez}-\operatorname{Imz})=4-(-3)+(0-(-7))=14$

[UPLIFTJEE]

Answer

14

[UPLIFTJEE]

Solution

Let $\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} 4 \mathrm{e}^{-\mathrm{i} \theta} \Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$
So locus of w is ellipse $\frac{x^{2}}{\left(\frac{17}{4}\right)^{2}}+\frac{y^{2}}{\left(\frac{15}{4}\right)^{2}}=1$ Locus of z is circle $x^{2}+y^{2}=16$ So intersect at 4 points

[UPLIFTJEE]

Answer

C

[UPLIFTJEE]

Solution

$\mathrm{w}_{1}=\mathrm{z}_{\mathrm{i}} \mathrm{i}=(5+4 \mathrm{i}) \mathrm{i}=-4+5 \mathrm{i}$

\[ W_{2}=z_{2}(-i)=(3+5 i)(-i)=5-4 i \]

\[ W_{1}-W_{2}=-9+8 i . \text { Principal argument }=\pi-\tan ^{-1}\left(\frac{8}{9}\right) \]

[UPLIFTJEE]

Answer

C

[UPLIFTJEE]

Solution

$(\mathrm{z}-2 \mathrm{i})(\overline{\mathrm{z}}+2 \mathrm{i})=4(\mathrm{z}+\mathrm{i})(\overline{\mathrm{z}}-\mathrm{i})$

\[ \begin{aligned} & z \bar{z}+4+2 i(z-\bar{z})=4(z \bar{z}+1+i(\bar{z}-z)) \\ & 3 z \bar{z}-6 i(z-\bar{z})=0 \Rightarrow x^{2}+y^{2}-2 i(2 i y)=0 \Rightarrow x^{2}+y^{2}+4 y=0 \end{aligned} \]

[UPLIFTJEE]

Answer

B

[UPLIFTJEE]

Solution

Let $\mathrm{z}=\mathrm{x}+\mathrm{iy},|\mathrm{z}+1|=1 \Rightarrow \sqrt{(\mathrm{x}-1)^{2}+\mathrm{y}^{2}}=1 \Rightarrow(\mathrm{x}-1)^{2}+\mathrm{y}^{2}=1$

\[ \begin{aligned} & (\sqrt{2}-1)(\mathrm{z}+\overline{\mathrm{z}})-\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})=2 \sqrt{2} \Rightarrow(\sqrt{2}-1)(2 \mathrm{x})-\mathrm{i}(2 \mathrm{iy})=2 \sqrt{2} \\ & \Rightarrow(\sqrt{2}-1) \mathrm{x}+\mathrm{y}=\sqrt{2} \Rightarrow \mathrm{y}=\sqrt{2}-(\sqrt{2}-1) \mathrm{x} \ldots \text { (ii) On solving with (i) } \\ & (\mathrm{x}-1)^{2}+2+(3-2 \sqrt{2}) \mathrm{x}^{2}-2 \sqrt{2}(\sqrt{2}-1) \mathrm{x}=1 \Rightarrow(2-\sqrt{2}) \mathrm{x}^{2}+(\sqrt{2}-3) \mathrm{x}+1=0 \text { For } \mathrm{x}=1, \mathrm{y}=1, \mathrm{z}=1+\mathrm{i} \text { for }|\mathrm{z}|=\sqrt{2} \text { (least modulus) } \end{aligned} \]

$\mathrm{x}=\frac{1}{2-\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}, \mathrm{y}=\frac{1}{\sqrt{2}}, \mathrm{z}=\frac{\sqrt{2}+1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}$, for $|\mathrm{z}|=\frac{\sqrt{4+2 \sqrt{2}}}{\sqrt{2}}=\sqrt{2+\sqrt{2}}$ (greatest modulus)
we get $\mathrm{z}_{1}\left(\frac{\sqrt{2}+1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$ and $\mathrm{z}_{2}(1+\mathrm{i})$ Now, $\left|\sqrt{2} \mathrm{z}_{1}-\mathrm{z}_{2}\right|^{2}=|\sqrt{2}+1+\mathrm{i}-(1+\mathrm{i})|^{2}=(\sqrt{2})^{2}=2$

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