# What is the value of $|\alpha|$?

Let the complex numbers $$\alpha$$ and $$\frac{1}{\bar{\alpha}}$$ (notice the bar above $$\alpha$$) lie on the circles $$(x-x_0)^2+(y-y_0)^2=r^2$$ and $$(x-x_0)^2+(y-y_0)^2=4r^2$$ respectively. If $$z_0=x_0+iy_0$$ satisfy $$2|z|^2=r^2+2$$. What is the value of $$|\alpha|$$?

I tried to represent the information given graphically and found that the two circles are concentric. Also the two given complex numbers lie on the same straight line since their arguments are the same. I do not understand how to proceed further. How do I solve this question?

If we write $$\;\alpha=a+ib\,,\,\;a,b\in\Bbb R\;$$ , then

$$\frac1{\overline\alpha}=\frac1{a-ib}=\frac{a+ib}{a^2+b^2}=\frac\alpha{a^2+b^2}$$

and then we get: $$\;\alpha\;$$ is on the first circle and $$\;\frac1{\overline\alpha}\;$$ is on the second one:

$$\begin{cases}(i)\;\,\;\;(a-x_0)^2+(b-y_0)^2=r^2\\{}\\(ii)\;\;\left(\cfrac a{a^2+b^2}-x_0\right)^2+\left(\cfrac b{a^2+b^2}-y_0\right)^2=4r^2\end{cases}$$

and we're also given that $$\;2(x_0^2+y_0^2)=r^2+2\;$$ , then we need quite a few substitutions and stuff:

$$(i)\;\;a^2+b^2-2(ax_0+by_0)+\overbrace{x_0^2+y_0^2}^{=\frac12r^2+1}=r^2\implies a^2+b^2-2(ax_0+by_0)=\frac12r^2-1$$

But also

$$(ii)\;\;\frac{a^2+b^2}{(a^2+b^2)^2}-2\left(\frac{ax_0+by_0}{a^2+b^2}\right)+\overbrace{x_0^2+y_0^2}^{=\frac12r^2+1}=4r^2\implies \frac{a^2+b^2}{(a^2+b^2)^2}-2\left(\frac{ax_0+by_0}{a^2+b^2}\right)=\frac72r^2-1$$

And from here we get that

$$\begin{cases}(i)\implies-2(ax_0+by_0)=\cfrac12r^2-1-a^2-b^2\\{}\\ (ii)\implies-2(ax_0+by_0)=\left[\cfrac72r^2-1- \cfrac{a^2+b^2}{(a^2+b^2)^2}\right](a^2+b^2)=\cfrac{7r^2-2}2(a^2+b^2)-1\end{cases}$$

and comparing right sides:

$$\left(\frac72r^2-1\right)(a^2+b^2)-1=\frac12r^2-1-a^2-b^2\implies\frac72r^2(a^2+b^2)=\frac12r^2\implies...$$

Finish now the argument.

$$|\alpha-z_0|=r\tag1$$ $$\big|\frac{1}{\bar{\alpha}}-z_0\big|=2r\tag2$$ and $$2|z_0|^2=r^2+2\tag 3$$ $$(1)$$ and $$(3)$$ give : $$(\alpha-z_0)(\bar{\alpha}-\bar{z_0})=r^2\Longrightarrow$$ $$-|\alpha|^2+{r^2\over 2}= 1-(\bar{\alpha}z_0+\bar{z_0}{\alpha}){\tag{1’}}$$ (2) gives $$\big(\frac{1}{\bar{\alpha}}-z_0\big)\big(\frac{1}{\alpha}-\bar{z_0}\big)=\frac{\overbrace{1-(\bar{\alpha}z_0+\bar{z_0}{\alpha})}^{-|\alpha|^2+\frac{r^2}{2}}+|\alpha|^2|z_0|^2 }{|\alpha|^2}=4r^2\tag{2’}$$ $$(1’)$$ and $$(2’)$$ give finally: $$4r^2|\alpha|^2=|\alpha|^2\underbrace{\big({r^2\over 2}+1\big)}_{|z_0|^2}-|\alpha|^2+{r^2\over 2}\iff 7|\alpha|^2=1\Longrightarrow |\alpha|=\frac{1}{\sqrt{7}}$$

Notation. To simplify my TEX job, let $$i$$ be the imaginary unit, $$\alpha := a + b i$$ and $$z := x + y i$$.

Note that $$\displaystyle\frac{1}{\bar \alpha} = \frac{a + b i}{a^2 + b^2} = \frac{\alpha}{\vert\alpha\vert^2}$$.

We thus have given: \begin{align} \tag 1 \label 1 \vert \alpha - z\vert^2 &= r^2, \\ \tag 2 \label 2 \left\vert \frac{\alpha}{\vert\alpha\vert^2} - z\right\vert^2 &= 4 r^2, \\ \tag 3 \label 3 \vert z\vert^2 &= \frac{r^2} 2 + 1. \end{align}

The left side of $$\eqref 1$$ equals $$a^2 - 2 a x + x^2 + b^2 - 2 b y + y^2$$ and thus $$$$\tag 4 \label 4 \vert\alpha\vert^2 - 2 a x - 2by + \vert z \vert^2 = r^2$$$$

By expanding and simplifying the left side of $$\eqref 2$$ (I'll leave this to you), one similarly gets $$$$\tag 5 \label 5 \frac 1{\vert\alpha\vert^2}(1 - 2 a x - 2 by) + \vert z\vert^2 = 4 r^2$$$$

By multiplying with $$\vert\alpha\vert^2$$, we can deduce from $$\ref 5$$ that $$$$\tag 6 \label 6 -2ax - 2by = \vert\alpha\vert^2(4 r^2 - \vert z\vert^2) -1$$$$

By using $$\eqref 6$$ in $$\eqref 4$$, we get $$$$\begin{split} &\vert\alpha\vert^2 (1 + 4r^2 - \vert z\vert^2) -1 + \vert z \vert^2 = r^2 \\ \iff & \vert\alpha\vert^2 = \frac{r^2+1-\vert z\vert^2}{1 + 4r^2 - \vert z\vert^2} \end{split} \tag 7 \label 7$$$$

By using $$\eqref 3$$ in $$\eqref 7$$, we deduce $$$$\tag{result} \vert\alpha\vert^2 = \frac{r^2+1-(\frac{r^2} 2 + 1)}{1 + 4r^2 - (\frac{r^2} 2 + 1)} = \frac{1}{7}.$$$$

• Typo in your last line. Apr 18, 2019 at 16:40
• @HAMIDINESOUMARE Where is the typo? :) Apr 18, 2019 at 16:53
• 1-1, there’s no 2. Apr 18, 2019 at 16:55
• @HAMIDINESOUMARE fixed now, thanks Apr 18, 2019 at 16:57