Let the complex numbers and 1 lie on the circles |z-z_0|^2=4 and |z-z_0|^2=16 respectively…

Question

Let the complex numbers $\alpha $ and $\frac{1}{\vec{\alpha}}$ lie on the circles ${|z-{z}_{0}|}^{2}=4$ and ${|z-{z}_{0}|}^{2}=16$ respectively, where ${z}_{0}=1+i$. Then, the value of $100|\alpha {|}^{2}$ is \_\_\_\_\_\_\_\_\_\_.

Answer: 20

Accepted Answer

$\begin{gathered} {|z-{z}_{0}|}^{2}=4 \ \Rightarrow (\alpha -{z}_{0})(\vec{\alpha}-{\vec{z}}_{0})=4 \ \Rightarrow \alpha \vec{\alpha}-\alpha {\vec{z}}_{0}-{z}_{0}\vec{\alpha}+{|{z}_{0}|}^{2}=4 \ \Rightarrow |\alpha {|}^{2}-\alpha {\vec{z}}_{0}-{z}_{0}\vec{\alpha}=2\, \ {|z-{z}_{0}|}^{2}=16 \end{gathered}$

$\begin{gathered} \Rightarrow (\frac{1}{\vec{\alpha}}-{z}_{0})(\frac{1}{\alpha }-{\vec{z}}_{0})=16 \ \Rightarrow (1-\vec{\alpha}{z}_{0})(1-\alpha {\vec{z}}_{0})=16|\alpha {|}^{2} \ \Rightarrow 1-\vec{\alpha}{z}_{0}-\alpha {\vec{z}}_{0}+|\alpha {|}^{2}{|{z}_{0}|}^{2}=16|\alpha {|}^{2} \ \Rightarrow 1-\vec{\alpha}{z}_{0}-\alpha {\vec{z}}_{0}=14|\alpha {|}^{2}\, \end{gathered}$

From (1) and (2)

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

Solutions · 1