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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.

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    $\begingroup$ You don't need polar coordinates for $z+1/z$, clearly. Then $\log z =\ln\sqrt(x^2+y^2)+i\arctan(y/x)$ $\endgroup$
    – saulspatz
    May 29 '21 at 16:14
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You don't need polar coordinates for $z+1/z$. Just substitute $z=x+iy$ and clear the denominator. Then $\log z =\ln\sqrt{x^2+y^2}+i\arctan(y/x)=\ln(x^2+y^2)/2+i\arctan(y/x)$

EDIT

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.

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  • $\begingroup$ 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. $\endgroup$
    – Seabourn
    May 29 '21 at 16:22

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