Solving an equation involving complex numbers.

Question

I tried solving the problem on my own. I would like to know if I have made any mistakes. If I have indeed made a mistake, I would appreciate it if someone corrects it and explains what it is. Also, I would like to know if the end argument about the real number is flawed. Thanks in advance.

Answer 1

Notice, when $z\in\mathbb{C}$:

$$\frac{\overline{z}}{z}=\frac{\overline{z}\overline{z}}{z\overline{z}}=\frac{\overline{z}^2}{|z|^2}=\frac{\left(\Re[z]-\Im[z]i\right)^2}{\Re^2[z]+\Im^2[z]}=\frac{\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i}{\Re^2[z]+\Im^2[z]}$$

So, we get that:

• $$\Re\left[\frac{\overline{z}}{z}\right]=\frac{\Re^2[z]-\Im^2[z]}{\Re^2[z]+\Im^2[z]}$$
• $$\Im\left[\frac{\overline{z}}{z}\right]=-\frac{2\Re[z]\Im[z]}{\Re^2[z]+\Im^2[z]}$$

Now, when $z=1+xi$, $\overline{z}=1-xi$ and $x\in\mathbb{R}$ we get:

$$\frac{1-xi}{1+xi}=\frac{1-x^2-2xi}{1+x^2}=\frac{1-x^2}{1+x^2}-\frac{2xi}{1+x^2}$$

If we now know that this have to equal $a-bi$ and $a\space\wedge b\space\in\mathbb{R}$, we get the system (given that $a^2+b^2=1$):

$$\frac{1-x^2}{1+x^2}-\frac{2xi}{1+x^2}=a-bi\Longleftrightarrow \begin{cases} \frac{1-x^2}{1+x^2}=a\\ -\frac{2x}{1+x^2}=-b\\ a^2+b^2=1 \end{cases} $$

Solving this, gives us (when $a+1\ne0$):

• $$b=\sqrt{1-a^2},x=\frac{\sqrt{1-a^2}}{a+1}$$
• $$b=-\sqrt{1-a^2},x=-\frac{\sqrt{1-a^2}}{a+1}$$