Complex Number Question 6
Question 6 - 2024 (27 Jan Shift 2)
Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $\left|z-z_{0}\right|^{2}=4$ and $\left|z-z_{0}\right|^{2}=16$ respectively, where $\mathrm{z}_{0}=1+\mathrm{i}$. Then, the value of $100|\alpha|^{2}$ is.
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Answer (20)
Solution
$\left|z-z_{0}\right|^{2}=4$
$\Rightarrow\left(\alpha-z_{0}\right)\left(\bar{\alpha}-\bar{z}_{0}\right)=4$
$\Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}{0}-z{0} \bar{\alpha}+\left|z_{0}\right|^{2}=4$
$\Rightarrow|\alpha|^{2}-\alpha \bar{z}{0}-z{0} \bar{\alpha}=2$
$\left|z-z_{0}\right|^{2}=16$
$\Rightarrow\left(\frac{1}{\bar{\alpha}}-z_{0}\right)\left(\frac{1}{\alpha}-\bar{z}_{0}\right)=16$
$\Rightarrow\left(1-\bar{\alpha} z_{0}\right)\left(1-\alpha \bar{z}_{0}\right)=16|\alpha|^{2}$
$\Rightarrow 1-\bar{\alpha} z_{0}-\alpha \bar{z}{0}+|\alpha|^{2}\left|z{0}\right|^{2}=16|\alpha|^{2}$
$\Rightarrow 1-\bar{\alpha} z_{0}-\alpha \bar{z}_{0}=14|\alpha|^{2}$
From (1) and (2)
$\Rightarrow 5|\alpha|^{2}=1$
$\Rightarrow 100|\alpha|^{2}=20$
