JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 29)

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 $$z_0=1+i$$. Then, the value of $$100|\alpha|^2$$ is __________.
Answer
20

Explanation

$$\begin{aligned} & \left|z-z_0\right|^2=4 \\ & \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 \\ & \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 \\ & \Rightarrow|\alpha|^2-\alpha \bar{z}_0-z_0 \bar{\alpha}=2 \quad\text{......... (1)} \\ & \left|z-z_0\right|^2=16 \end{aligned}$$

$$\begin{aligned} & \Rightarrow\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=16 \\ & \Rightarrow\left(1-\bar{\alpha} z_0\right)\left(1-\alpha \bar{z}_0\right)=16|\alpha|^2 \\ & \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0+|\alpha|^2\left|z_0\right|^2=16|\alpha|^2 \\ & \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0=14|\alpha|^2 \quad\text{....... (2)} \end{aligned}$$

From (1) and (2)

$$\begin{aligned} & \Rightarrow 5|\alpha|^2=1 \\ & \Rightarrow 100|\alpha|^2=20 \end{aligned}$$

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