I am note sure about an exercise which asks to find a right inverse function of $w=z^3$
May I simply answer that is $z=\sqrt[3]{|w|}(\cos\frac{\theta+2k\pi}{3}+i\sin\frac{\theta+2k\pi}{3})$?
In the same question, I have to find a right inverse function g for $w=z^4$ such that $g(-1)=\frac{\sqrt(2)(1+i)}{2}$... I know that If I choose $k=0$ and $\theta =\pi$ in the n-th root formula, I reach that condition. So, should the answer be $g(w)=\sqrt[4]{|w|}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$?
On
You can find a "Cartesian" formula for the fourth root.
Let $z^2=(u+iv)^2=x+iy$. By identification,
$$u^2-v^2=x,\\2uv=y$$
so that
$$(u^2-v^2)^2+4u^2v^2=(u^2+v^2)^2=x^2+y^2$$
and the positive solutions are
$$u^2=\frac{\sqrt{x^2+y^2}+x}2,\\v^2=\frac{\sqrt{x^2+y^2}-x}2.$$
So we have established
$$\sqrt{x+iy}=\pm\frac12\left(\sqrt{\sqrt{x^2+y^2}+x}+i\sqrt{\sqrt{x^2+y^2}-x}\right)$$
and it suffices to iterate to get
$$\sqrt{\sqrt{x+iy}}= \\\pm\frac12\left(\sqrt{\sqrt{2(x^2+y^2)}\pm\frac12\sqrt{\sqrt{x^2+y^2}+x}}+i\sqrt{\sqrt{2(x^2+y^2)}\mp\frac12\sqrt{\sqrt{x^2+y^2}+x}}\right).$$
Note that the inner $\pm\,\mp$ are "synchronized" and there are four solutions.
Your answers are on the right track. For the $z^3$ problem you need to specify that $\theta = \operatorname{Arg}(w)$, and state what $k$ is (an arbitrary integer in your case). For $z^4$ you can't just take $\theta = \pi$ because $\theta = \operatorname{Arg}(w)$ varies with $w$. (The $g(w)$ you wrote has $g(w)^4 = -|w|$ which is not what you want). You can use a similar approach to what you did on $z^3$ be defining $$ g(w) = \sqrt[4]{|w|} \left(\cos\left(\frac{\operatorname{Arg}(w) + 2 k \pi}{4}\right) + \sin\left(\frac{\operatorname{Arg}(w) + 2 k \pi}{4}\right)\right) $$ Then you need to determine what value of $k$ will have $g(-1) = \frac{\sqrt2}2(1 + i)$