Squareroot of complex number to the square $\sqrt{z^2}$

Question

I have to calculate $\sqrt{z^2}$ an I am confused about how to procede.

I thought about introducing $z=|z|\exp(i\phi+2\pi k) \implies z^2=|z|^2\exp(2i\phi+4\pi k)$. Hence,

$$\sqrt{z^2}=\sqrt{|z|^2\exp(2i\phi+4\pi k)}=|z|\exp(i\phi+2\pi k)=|z|\exp(i\phi)=z.$$

Is that right? That seems strange to me, but i don't see any mistake in my calculations.

Answer 1

When talking about complex numbers theres a thing called branch which help us consider how to deal with a multivalued functions.

$\sqrt{z}$ is a multivalued function (you surely know that $\sqrt{4}=\pm2$). in order to have a spcific value for $\sqrt{z^2}$ you need to know in which branch you are. once you'll have to answer to that, you could answer your question. basiclly, the two branches will be $f(z)=z$ or $f(z)=-z$.

Edit:

The mistake in your calculation is the following:

$\sqrt{z^2}=\sqrt{|z|^2\exp(2i\phi+4\pi k)}=\sqrt{|z|^2}\sqrt{\exp(2i\phi+4\pi k)}=\pm|z|\exp(i\phi+2\pi k)=\pm|z|\exp(i\phi)=\pm z$

Hope this answer helps!

Answer 2

Your answer is correct. The point of the question is to make you think about what a complex square root is. As you might know, it is not possible to define a holomorphic or even continuous inverse of $z \mapsto z^2$ on $\mathbb{C}$, as you have to possible choices for an inverse. For the square the both possibilities only differ by a minus sign. I suggest you read about square roots and branches of square roots.