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I have a task which I do not understand:

Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.

Given that Im $w = 0$, show that $| z | = 1$.

Partial solution (thanks to @ABC and @aranya):

If I substitute $z$ with $x + iy$ then we have $w = \frac{x + iy}{x^2+2xyi-y^2+1}$ or written slightly different $w= \frac{x + iy}{x^2-y^2+1+2xyi}$ or $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}$.

In this format we can multiply with complex conjugate $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}\cdot\frac{(x^2-y^2+1)-(2xyi)}{(x^2-y^2+1)-(2xyi)}$.

Then we get $w=\frac{x^3+xy^2-x^2yi-y^3i+x+iy}{(x^2+^2+1)^2-(2xyi)^2} = \frac{(x^3+xy^2+x)+(-x^2yi-y^3i+iy)}{(x^2+^2+1)^2+(2xy)^2} = \frac{(x^3+xy^2+x)+i(-x^2y-y^3+y)}{(x^2+^2+1)^2+(2xy)^2}$.

And as stated above, imaginary part of $w = 0$ meaning $\frac{-x^2y-y^3+y}{(x^2+^2+1)^2+(2xy)^2} = 0$.

As denominator can't be zero, means nominator is zero $-x^2y-y^3+y=0$. Or $y\cdot(-x^2-y^2+1)=0$. As $y\not=0$ implies $-x^2-y^2+1=0$ or $x^2+y^2=1$.

What now?

share|improve this question
    
You have a relation $w=\frac{z}{z^2+1}$ that for each number $z$ gives you a number $w$. You are supposed to assume that the imaginary part of $w$ is zero, i.e. $w$ is a real number. You have to show that the number $z$ you put in the right hand side has absolute value equal to one. –  ABC Nov 11 '13 at 19:33
    
hint: if we expand using x and y and then multiply by complex conjugate of the denominator then what do we get? –  aranya Nov 11 '13 at 19:33

1 Answer 1

$$w=\frac z{z^2+1}=\frac{|z|^2\overline z+z}{|z^2+1|^2}$$

and thus

$$\text{Im}\,w=0\iff \text{Im}\,z\cdot\left(|z|^2-1\right)=0\stackrel{\text{since Im}\,z=y\neq 0}\iff |z|^2=1\;\ldots$$

share|improve this answer
    
I just started learning complex numbers and I do not fully follow you. –  Anja_7 Nov 11 '13 at 20:29
    
@Anja_7, you just need to be sure you know or can prove that $$\forall\,z=x+iy\in\Bbb C\;\;,\;\;z\overline z=|z|^2\;,\;\;|z|^2\overline z+z=|z|^2(x-iy)+x+iy$$and thus the imaginary part of $\;w\;$ is what is written in the answer above. –  DonAntonio Nov 11 '13 at 20:40

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