# Thread: Complex roots - Equation

1. ## Complex roots - Equation

Show that the real part of any solution of $(z + 100)^{100} = (z - 1)^{100}$ must be 0.

Now in the book they solve it by starting: $(z + 100)^{100} = (z - 1)^{100} \Rightarrow (z +1) = (z - 1)e^{\frac{2\pi ki}{100}}$.

I understand the steps that follow it, however I don't understand that step. I referred to the definition of m-th roots of complex numbers, which says: $z^{1/m} = |z|^{1/m}e^{i(\theta + 2k\pi)/m}$

What am I missing? Thanks in advance.

2. Hello,

Recall that $e^{2ik\pi}=1$

We have :
$(z + 100)^{100} = (z - 1)^{100} e^{2ik\pi}$

Take the 100-th root on both side to get :
$(z+100)=(z-1) e^{\frac{2ik\pi}{100}}$

What I don't understand is why they wrote $(z+{\color{red}1})=\dots$

3. Originally Posted by Moo
Hello,

Recall that $e^{2ik\pi}=1$

We have :
$(z + 100)^{100} = (z - 1)^{100} e^{2ik\pi}$

Take the 100-th root on both side to get :
$(z+100)=(z-1) e^{\frac{2ik\pi}{100}}$

What I don't understand is why they wrote $(z+{\color{red}1})=\dots$
Hi,
Yes, thanks for the tip, I see it now. That was a typo on my part it should say $(z + 1)^{100}$, where I accidentally typed $(z + 100)^{100}$, then I copy-pasted the typo.

Thanks again.

Now I'm having trouble understand a subsequent step, which I thought I understood.
They get all the way to $z = \frac{e^{2\pi k i/100} + 1}{e^{2\pi k i/100} - 1} = \frac{e^{\pi k i/100} + e^{-\pi k i/100}}{e^{\pi k i/100} - e^{\pi k i/100}} = -i\frac{\cos{(\pi k/100)}}{\sin{(\pi k/100)}}$. To me the it seems the denominator is 0 in the middle step, that's where my problem is, as I don't see how they made that substitution either.