Topic 2 Complex Number JEE MAIN 2024
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Let $\mathrm{S}=\{\mathrm{z} \in \mathrm{C},|\mathrm{z}-1|=1$ and $(\sqrt{2}-1)(\mathrm{z}+\overline{\mathrm{z}})-\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})=2 \sqrt{2}\}$. Let $\mathrm{z}_{1}, \mathrm{z}_{2} \in \mathrm{~S}$ be such that $\left|\mathrm{z}_{1}\right|=\max _{\mathrm{Z} \in \mathrm{S}}|\mathrm{z}|$ and $\left|\mathrm{z}_{2}\right|=\min _{\mathrm{Z} \in \mathrm{S}}|\mathrm{z}|$. Then $\left|\sqrt{2} \mathrm{z}_{1}-\mathrm{z}_{2}\right|^{2}$ equals
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Consider the following two statements : Statement I : For any two non-zero complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2},\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right| \leq 2\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)$,
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Let z be a complex number such that the real part of $\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}$ is zero. Then, the maximum value of $|\mathrm{z}-(6+8 \mathrm{i})|$ is equal to
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Let $\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}| \leq 5\}, \mathrm{S}_{2}=\left\{\mathrm{z} \in \mathrm{C}: \operatorname{Im}\left(\frac{\mathrm{z}+1-\sqrt{3} \mathrm{i}}{1-\sqrt{3} \mathrm{i}}\right) \geq 0\right\}$ and $\mathrm{S}_{3}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{z}) \geq 0\}$. Then the area of the region $\mathrm{S}_{1} \cap \mathrm{~S}_{2} \cap \mathrm{~S}_{3}$ is
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If $\mathrm{z}_{1}, \mathrm{z}_{2}$ are two distinct complex number such that $\left|\frac{\mathrm{z}_{1}-2 \mathrm{z}_{2}}{\frac{1}{2}-\mathrm{z}_{1} \overline{\mathrm{z}}_{2}}\right|=2$, then
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The sum of the square of the modulus of the elements in the set $\{\mathrm{z}=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathrm{Z}, \mathrm{z} \in \mathrm{C},|\mathrm{z}-1| \leq 1,|\mathrm{z}-5| \leq|\mathrm{z}-5 \mathrm{i}|\}$ is \_\_\_\_.
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Let $\mathrm{P}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}+2-3 \mathrm{i}| \leq 1\}$ and $\mathrm{Q}=\{\mathrm{z} \in \mathrm{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \leq-8\}$. Let in $\mathrm{P} \cap \mathrm{Q},|\mathrm{z}-3+2 \mathrm{i}|$ be maximum and minimum at $z_{1}$ and $z_{2}$ respectively. If $\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals
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The area (in sq. units) of the region $\mathrm{S}=\{\mathrm{z} \in \mathrm{C} ;|\mathrm{z}-1| \leq 2 ;(\mathrm{z}+\overline{\mathrm{z}})+\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}}) \leq 2, \operatorname{Im}(\mathrm{z}) \geq 0\}$ is
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If z is a complex number such that $|\mathrm{z}| \geq 1$, then the minimum value of $\left|\mathrm{z}+\frac{1}{2}(3+4 \mathrm{i})\right|$ is
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Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $\left|\mathrm{z}-\mathrm{z}_{0}\right|^{2}=4$ and $\left|\mathrm{z}-\mathrm{z}_{0}\right|^{2}=16$ respectively, where $\mathrm{z}_{0}=1+\mathrm{i}$. Then, the value of $100|\alpha|^{2}$ is
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If $\alpha$ denotes the number of solutions of $|1-\mathrm{i}|^{\mathrm{x}}=2^{\mathrm{x}}$ and $\beta=\left(\frac{|\mathrm{z}|}{\arg (\mathrm{z})}\right)$, where $\mathrm{z}=\frac{\pi}{4}(1+\mathrm{i})^{4}\left[\frac{1-\sqrt{\pi \mathrm{i}}}{\sqrt{\pi}+\mathrm{i}}+\frac{\sqrt{\pi}-\mathrm{i}}{1+\sqrt{\pi \mathrm{i}}}\right], \mathrm{i}=\sqrt{-1}$ then the distance of the point $(\alpha, \beta)$ from the line $4 \mathrm{x}-3 \mathrm{y}=7$ is
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Let $\mathrm{S}=\{\mathrm{z} \in \mathrm{C},|\mathrm{z}-1|=1$ and $(\sqrt{2}-1)(\mathrm{z}+\overline{\mathrm{z}})-\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})=2 \sqrt{2}\}$. Let $\mathrm{z}_{1}, \mathrm{z}_{2} \in \mathrm{~S}$ be such that $\left|\mathrm{z}_{1}\right|=\max _{\mathrm{Z} \in \mathrm{S}}|\mathrm{z}|$ and $\left|\mathrm{z}_{2}\right|=\min _{\mathrm{Z} \in \mathrm{S}}|\mathrm{z}|$. Then $\left|\sqrt{2} \mathrm{z}_{1}-\mathrm{z}_{2}\right|^{2}$ equals
