# Transforming a complex function into $F(z) = u(x,y)+iv(x,y)$

I have the following function: $$\begin{equation*} F(z) = \frac{1}{2}\left(z+\frac{1}{z}\right)+ik\ln(z), \quad k \geq 0 \end{equation*}$$ where $$\ln$$ is the main branch of the complex logarithm.

And I need to transform it into the form: $$\begin{equation*} F(z) = u(x,y) + iv(x,y) \end{equation*}$$ What I have done. My idea was to use polar coordinates aproach to the function, with $$z=re^{i\theta}$$, where $$r=|z|$$ and $$\theta \in Arg(z)$$. After a few calculations I've got the following result: $$\begin{equation*} F(z) = u(r,\theta)+iv(r,\theta) = \left[\frac{\cos\theta}{2}\left(r+\frac{1}{r}\right)-k\theta\right]+i\left[\frac{\sin\theta}{2}\left(r-\frac{1}{r}\right)+k\ln(r)\right] \end{equation*}$$

Is there an aproach to calculate it in terms of $$x$$ and $$y$$ without being "reverting" the polar coordinates? Because I understand that $$\theta = tg^{-1}(\frac{y}{x})$$ and $$r=|z|=\sqrt{x^2+y^2}$$ making me able to obtain a relation like I wanted but i was wondering if there's a way to do it without using polar coordinates at first.

Thanks for all the help is advance.

• You don't need polar coordinates for $z+1/z$, clearly. Then $\log z =\ln\sqrt(x^2+y^2)+i\arctan(y/x)$ May 29 '21 at 16:14

You don't need polar coordinates for $$z+1/z$$. Just substitute $$z=x+iy$$ and clear the denominator. Then $$\log z =\ln\sqrt{x^2+y^2}+i\arctan(y/x)=\ln(x^2+y^2)/2+i\arctan(y/x)$$
In response to OP's comment, yes that's correct. An easy way to check this is that when $$|z|=1$$, we have $$\frac1z=\overline{z}$$, so that $$z+\frac1z=z+\overline{z}=2\operatorname{Re} z$$. Then the real part of $$\log z$$ is $$0$$, so $$i\log z$$ is also real.
• That would make the following: \begin{equation*} v(x,y) = \frac{-y+y^3+x^2y}{2x^2+2y^2} + \frac{k}{2}\ln(x^2+y^2) \end{equation*} So, now let's say I wanted to calculate $v(x,y)$ for $|z| = 1 \Leftrightarrow x^2 = 1 - y^2$. I would have: \begin{equation*} v(x,y) = 0 + \frac{k}{2}ln(1) = 0 + 0 = 0 \end{equation*}. Is this correct indeed? Thanks for your help. May 29 '21 at 16:22