Determine the number of point $z \in \mathbb{C}$ such that $(z-2)^4=(1+i)^4$

Question

I'm new in Real Analysis, and I'm stuck as I would like to determine the number of point $z \in \mathbb{C}$ such that $(z-2)^4=(1+i)^4$.

I might transform $z$ in $re^{i\alpha}$, but I don't know how to proceed. Can someone help me explaining in detail the steps? Thank you very much!

Answer 1

HINT

We have that

• $w=(1+i)^4$

then evaluate

$$(z-2)^4=w$$

Answer 2

$$(z-2/1+i )^4=1=e^{2n\pi i}$$ where $n$ is any integer

$(z-2)/(1+i)=e^{2n\pi i /4}$ where $n=0,1,2,3$

Answer 3

Basically: $$(1+i)^4=(z-2)^4 \to z-2=\pm 1 \pm i $$

Should be easy from there

Answer 4

You said you want the number of points, not their exact location.

For you may note that the original equation is a polynomial equation $P(z)=0$ of degree $4$.

The derivative $P^\prime(z)$ is $4(z-2)^3$ which has $z=2$ has the only zero with multiplicity $3$. Since $z=2$ is not a root of $P(z)$ the original equation has $4$ different solutions.