Complex Numbers 2 Question 8
8. Let complex numbers $\alpha$ and $1 / \bar{\alpha}$ lies on circles $\left(x-x _0\right)^{2}+\left(y-y _0\right)^{2}=r^{2}$ and $\left(x-x _0\right)^{2}+\left(y-y _0\right)^{2}=4 r^{2}$, respectively.
If $z _0=x _0+i y _0$ satisfies the equation $2\left|z _0\right|^{2}=r^{2}+2$, then $|\alpha|$ is equal to
(2013 Adv.)
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) $\frac{1}{\sqrt{7}}$
(d) $\frac{1}{3}$
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Answer:
Correct Answer: 8. (c)
Solution:
- PLAN Intersection of circles, the basic concept is to solve the equations simultaneously and using properties of modulus of complex numbers.
Formula used $\quad|z|^{2}=z \cdot \bar{z}$
and
$$ \begin{aligned} \left|z _1-z _2\right|^{2} & =\left(z _1-z _2\right)\left(\bar{z} _1-\bar{z} _2\right) \\ & =\left|z _1\right|^{2}-z _1 \bar{z} _2-z _2 \bar{z} _1+\left|z _2\right|^{2} \end{aligned} $$
Here, $\left(x-x _0\right)^{2}+\left(y-y _0\right)^{2}=r^{2}$
and $\left(x-x _0\right)^{2}+\left(y-y _0\right)^{2}=4 r^{2}$ can be written as,
$$ \left|z-z _0\right|^{2}=r^{2} \text { and }\left|z-z _0\right|^{2}=4 r^{2} $$
Since, $\alpha$ and $\frac{1}{\bar{\alpha}}$ lies on first and second respectively.
$\therefore \quad\left|\alpha-z _0\right|^{2}=r^{2}$ and $\left|\frac{1}{\bar{\alpha}}-z _0\right|^{2}=4 r^{2}$
$\Rightarrow \quad\left(\alpha-z _0\right)\left(\bar{\alpha}-\bar{z} _0\right)=r^{2}$
$\Rightarrow \quad|\alpha|^{2}-z _0 \bar{\alpha}-\bar{z} _0 \alpha+\left|z _0\right|^{2}=r^{2}$
and $\quad\left|\frac{1}{\bar{\alpha}}-z _0\right|^{2}=4 r^{2}$
$\Rightarrow \quad \frac{1}{\bar{\alpha}}-z _0 \quad \frac{1}{\alpha}-\bar{z} _0=4 r^{2}$
$\Rightarrow \quad \frac{1}{|\alpha|^{2}}-\frac{z _0}{\alpha}-\frac{\bar{z} _0}{\bar{\alpha}}+\left|z _0\right|^{2}=4 r^{2}$
Since, $\quad|\alpha|^{2}=\alpha \cdot \alpha$
$\Rightarrow \frac{1}{|\alpha|^{2}}-\frac{z _0 \cdot \bar{\alpha}}{|\alpha|^{2}}-\frac{\bar{z} _0}{|\alpha|^{2}} \cdot \alpha+\left|z _0\right|^{2}=4 r^{2}$
$\Rightarrow \quad 1-z _0 \bar{\alpha}-\bar{z} _0 \alpha+|\alpha|^{2}\left|z _0\right|^{2}=4 r^{2}|\alpha|^{2}$
On subtracting Eqs. (i) and (ii), we get
$$ \begin{array}{rlrl} \left(|\alpha|^{2}-1\right)+\left|z _0\right|^{2}\left(1-|\alpha|^{2}\right) & =r^{2}\left(1-4|\alpha|^{2}\right) \\ \Rightarrow \quad\left(|\alpha|^{2}-1\right)\left(1-\left|z _0\right|^{2}\right) & =r^{2}\left(1-4|\alpha|^{2}\right) \\ \Rightarrow \quad\left(|\alpha|^{2}-1\right) 1-\frac{r^{2}+2}{2} & =r^{2}\left(1-4|\alpha|^{2}\right) \\ & \text { Given, } & \left|z _0\right|^{2} & =\frac{r^{2}+2}{2} \end{array} $$
$$ \begin{aligned} & \Rightarrow \quad\left(|\alpha|^{2}-1\right) \cdot \frac{r^{2}}{2}=r^{2}\left(1-4|\alpha|^{2}\right) \\ & \Rightarrow \quad|\alpha|^{2}-1=-2+8|\alpha|^{2} \\ & \Rightarrow \quad 7|\alpha|^{2}=1 \\ & \therefore \quad|\alpha|=1 / \sqrt{7} \end{aligned} $$
