I am studying for an algebra final exam and I am stuck with this exercise:

Find all the $z\in\mathbb{C}$ that satisfy $(z+1)^4=(z+i)^2$

We have been shown a couple of similar examples in class, but none of them work with this equation. The ones we studied are much nicer. For instance, the complex numbers inside the parenthesis are not the same (one is $(z+1)$ while the other is $(z+i)$ ) and no solutions are easy to find to make things simpler. I mention this because in previous exercises those observations worked.

I tried a couple of things: I noticed that they do share their absolute value, and thought of using that to find the argument, with no luck.

After that, I tried finding solutions and writing a different factorization but could not find a way to smooth the terms of the equation. So, I tried the crude path of distributing the powers, but I have the feeling that there must be a more clever way of solving this. Also, even this path does not seem to be working.

Maybe there is a trick we haven't studied or something simpler that I am just too tired to notice. In any case, I'd welcome some help.


2 Answers 2


You have $u^2=v^2$ where $u=(z+1)^2$ and $v=z+i$. Surely $u=v$ or $u=-v$?


note that $$(z+1)^4-(z+i)^2$$ can factorized as $$(z-i) (z+(1+i)) \left(z^2+3 z+(1+i)\right)$$


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