Question

Complex numbers problem

Answer 1

\[\arg(\frac{ z+i }{ z-i })=\frac{ \pi }{ 4 }\]

Answer 2

So you want to find the complex number \(z\) such that the argument of \(\dfrac{z+i}{z-i}\) is \(\dfrac{\pi}{4}\) ?

Answer 3

Letting \(w=\dfrac{z+i}{z-i}=re^{i\pi/4}\), we get that
\[\begin{cases}r=\left|w\right|\\\\
\tan \dfrac{\pi}{4}=1=\dfrac{\text{Im}\left(w\right)}{\text{Re}\left(w\right)}~~\iff~~\text{Im}(w)=\text{Re}(w)\end{cases}\]
Let \(z=a+bi\). Then you have
\[\begin{align*}
w&=\frac{a+bi+i}{a+bi-i}\\\\
&=\frac{a+(b+1)i}{a+(b-1)i}\cdot\frac{a-(b-1)i}{a-(b-1)i}\\\\
&=\frac{a^2+a(b+1)i-a(b-1)i+(b+1)(b-1)}{a^2+(b-1)^2}\\\\
&=\frac{a^2+b^2-1+2ai}{a^2+b^2-2b+1}\\\\
&=\frac{a^2+b^2-1}{a^2+b^2-2b+1}+\frac{2a}{a^2+b^2-2b+1}i
\end{align*}\]
Since the real and imaginary parts are equal, you have
\[\begin{align*}a^2+b^2-1&=2a\\\\
a^2-2a+b^2&=1\\\\
a^2-2a+1+b^2&=2\\\\
(a-1)^2+b^2&=2
\end{align*}\]

Answer 4

Thank you for that explanation!

Answer 5

It would appear that there's an infinite number of solutions for the appropriate \(a\) and \(b\), but I would suggest that you take a few that satisfy the circle equation and see if they actually work with the initial equation.