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I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that $$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$

has only real solutions

This is what I got till now.

$$\left(\frac{1+i(a+ib)}{1-i(a+ib)}\right)^n = a+ib$$ $$\left(\frac{(1-b)+ia}{(1+b)-ia}\right)^n = a+ib$$

$$\frac{(1-b)+ia}{(1+b)-ia}.\frac{(1+b)+ia}{(1+b)+ia} = \frac{1-b^2+2ia-a^2}{1+2b+b^2+a^2}$$

then

$$\left(\frac{1-b^2-a^2}{1+2b+b^2+a^2}+\frac{2ia}{1+2b+b^2+a^2}\right)^n = a+ib$$ where $$a=0 \text{; & } 1+2b+b^2\neq0$$

Ro_Mac
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  • You are assuming that $z = a + bi$. In general, $z$ the solution to your equation may be completely different from the constant term $a + bi$ on the RHS. – balddraz Jul 29 '19 at 03:30
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    Welcome to MSE. I believe the question is asking for the conditions on $a$ and $b$ so the only complex values of $z$ which are solutions of the equality are those where $z$ is a real number, i.e., it has a $0$ imaginary part. – John Omielan Jul 29 '19 at 03:30
  • Thanks @JohnOmielan , yes I'm looking for the condition on a and b so the only complex values of z which satisfy the equality are those where z is a real number. But I don't know if I'm right on what I got till now. – Ro_Mac Jul 29 '19 at 03:38
  • @Ro_Mac As ZeroXLR stated, there's no direct connection, or other reason to believe, that $z = a + ib$ is a solution. Thus, you should change your solution technique to, instead, see what the LHS will generally be when $z$ is just a real number (e.g., by converting the numerator & denominator to polar form, take the $n$'th power, and then convert back to a complex number form). – John Omielan Jul 29 '19 at 03:42
  • Note that in the first step, you multiplied by (1+i)/(1-i) on the Left Hand Side only. This is not correct. – NoChance Jul 29 '19 at 14:01

3 Answers3

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Note that $$ |1-iz|^2 - |1+iz|^2 = -2i(z -\bar z) = 4 \operatorname{Im}(z) $$ so that $$ z \in \Bbb R \iff \left | \frac{1+iz}{1-iz}\right|= 1 \, . $$ It follows that $$ \left(\frac{1+iz}{1-iz}\right)^n = a+ib $$ has only real solutions $z$ if and only if $|a+ib|=1$, i.e. if $a^2+b^2=1$.

Martin R
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  • why do you substract the modules? – Ro_Mac Jul 29 '19 at 05:48
  • @Ro_Mac Because $\left | \frac{1+iz}{1-iz}\right|= 1$ is equivalent to $|1-iz|^2 - |1+iz|^2 = 0$. – Martin R Jul 29 '19 at 06:20
  • sorry for my ignorance, but why is $\left | \frac{1+iz}{1-iz}\right|= 1$, are you assuming it? That means $z=0$? And why did you use modules? – Ro_Mac Jul 29 '19 at 06:38
  • @Ro_Mac: I am not assuming anything. I show that $z \in \Bbb R$ is equivalent to $\left | \frac{1+iz}{1-iz}\right|= 1$, and therefore your equation has real solutions if and only if $a^2+b^2=1$. – Martin R Jul 29 '19 at 06:40
  • So that means $z = 0$? – Ro_Mac Jul 29 '19 at 06:44
  • @Ro_Mac: No. You asked for a condition on $a, b$ that your equation has only real solutions $z$. That condition is $a^2+b^2=1$. – Martin R Jul 29 '19 at 06:46
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that is an awesome answer given by Martin R. alternatively, let $$ a+ib = \sqrt{a^2+b^2} e^{i \theta} $$ $c$ be the real solution for satisfied $a,b$. One would obtain $$ \label{eq} \sqrt[n]{a^2+b^2} e^{ \frac{i (\theta + 2k\pi )}{n} } = \frac{1-c^2 +i 2c}{1+c^2} \tag1 $$ note that $$ ( \frac{1-c^2 }{1+c^2} ) ^2 + ( \frac{2c}{1+c^2} ) ^2 = 1 $$ which means $\sqrt[n]{a^2+b^2} =1$ equivalently $$a^2+b^2 =1 $$ And for each $e^{ \frac{i (\theta + 2k\pi )}{n} } $ there exists only one $c \in \mathbb{R}$ satisfy $\eqref{eq}$

bruceyuan
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Set $$ w=\frac{1+iz}{1-iz} $$ so you can solve for $z$, getting $$ z=i\frac{1-w}{1+w} $$ This is real if and only if $$ i\frac{1-w}{1+w}=-i\frac{1-\bar{w}}{1+\bar{w}} $$ that becomes $$ 1+\bar{w}-w-w\bar{w}=-1+\bar{w}-w+w\bar{w} $$ that is, $w\bar{w}=1$.

Therefore $|a+bi|=1$ is a necessary condition. Now if we write $a+bi$, the equation becomes $$ \frac{1+iz}{1-iz}=u $$ where $u$ is any $n$-th root of $a+bi$ and $|u|=1$. By the same argument as before, the solution is real.

egreg
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