Prove $\ \frac{z^{2010} - \bar z^{2010}}{1+z\bar z}$ is imaginary number

Question

Prove $\ \frac{z^{2010} - \bar z^{2010}}{1+z\bar z}$ is imaginary number.

I understand that if $\ z = (a+bi) $ then $\ z - \bar z = 2bi $ and the denominator $\ 1+z\bar z $ is $\ 1+|z|^2 $ and therefore it is a real number. so need to prove the numerator is imaginary.

I first tried to see what happens if I take $\ z^2 - \bar z^2 $ and it is imaginary $\ (a+bi)^2-(a-bi)^2 = 4abi $

and also $ z^{2010} = (z^{1005})^2$ so if $\ z^{1005} $ is imaginary number... but I have no clue what is $\ z^{1005}$ and maybe it's a real number..?

Answer 1

It’s much simpler than it might look at first glance.

First step: No matter what the (nonreal) complex number $z=a+bi$ is, $z-\bar z$ is purely imaginary, namely equal to $2bi$.

Second step: $(\bar z)^m=\overline{(z^m)}$.

Third step: no matter what the complex number $z$ is, we get $z\bar z$ to be real and nonnegative, equal to $a^2+b^2$.

Put them all together and see that your fraction is a purely imaginary number divided by a positive real number.

Answer 2

Hint:

$$\overline{z}^{2010}=\overline{(z^{2010})}$$

Also, $$\mathfrak I(w)=\frac{w-\bar w}{2i}$$ as you have noted.

Answer 3

We have that

$$\bar w=\overline{\ \frac{z^{2010} - \bar z^{2010}}{1+z\bar z}}=\ \frac{\bar z^{2010} - z^{2010}}{1+\bar zz}=-\ \frac{z^{2010} - \bar z^{2010}}{1+z\bar z}=-w$$

Answer 4

$z\bar z=|z|^2$

$(a+ib)^n-(a-ib)^n=2i\sum_{r=0}^{2r+1\le n}\binom n{2r+1}a^{n-(2r+1)}b^{2r+1}(-1)^r$

Answer 5

Using trigonometric form: $$z^{2010}-\bar{z}^{2010}=r^{2010}(\cos (2010t)+i\sin (2010t))-r^{2010}(\cos(2010t)-i\sin(2010t))=\\ 2r^{2010}\sin(2010t)\cdot i.$$

Answer 6

1) Note $1+z\overline{z} >0$, since $|z|^2=z\overline{z} \ge 0.$

Suffices to show that $z^{2010} - \overline{z}^{2010}$ is imaginary.

2) Set $z=re^{i\phi}$, then $\overline{z}=re^{-i\phi}$.

$z^{2010}=Re^{i\alpha}$ , and $\overline{z}^{2010}=Re^{-i\alpha}$,

where $R=r^{2010}$, and $\alpha=2010 \phi$.

$z^{2010}-\overline{z}^{2010}=$

$R\cos \alpha +iR\sin \alpha$

$- R \cos (-\alpha) -iR\sin(-\alpha)=$

$i2R\sin (\alpha)$, imaginary.