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Show that $F(z)=\frac{i}{2}\log(z+i)-\frac{i}{2}\log(z-i)$ is an antiderivative of $f(z)=\frac{1}{z^2+1}$ for $\operatorname{Re}(z)>0$. Is $F(z)$ equal to $\arctan z$?

I basically did $$\frac{d}{dz}F(z)=\frac{i}{2}\frac{1}{z+i}-\frac{i}{2}\frac{1}{z-i}=\frac{1}{z^2+1}=f(z)$$ Thus, $F(z)$ is an antiderivative of $f(z)$.

But my question is :

  1. Why do we need $\operatorname{Re}(z)>0$ here? Is that a preparation for $\arctan z$ to be defined or something?

  2. I think even though $F'(z)=(\arctan z)'=f(z)$, we cannot say $F(z)$ equal to $\arctan z$. Since they are only one member of the antiderivative family of $\frac{1}{z^2+1}$. Is that correct?

Thanks~

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    $\begingroup$ en.wikipedia.org/wiki/Atan2#Definition_and_computation $\endgroup$ – lab bhattacharjee Jan 9 '18 at 18:27
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    $\begingroup$ $\Re(z) > 0$ is to ensure the log are differentiable. If you defined the branch cut of $\ln z$ at the negative $x$-axis, then $\ln (z\pm i)$ are analytic at $\Re(z) > 0$, it cannot be extended to $\Re (z) > c$ for negative $c$. $\endgroup$ – pisco Jan 9 '18 at 18:34
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$\Re(z) > 0$ is to ensure the log are differentiable. If you defined the branch cut of $\ln z$ at the negative $x$-axis, then $\ln (z\pm i)$ are analytic at $\Re(z) > 0$, it cannot be extended to $\Re (z) > c$ for negative $c$.


I assume your $\ln z$ takes principal value. $F(z)$ does not equal to $\arctan z$. In fact, some simple calculation gives $$F(1) = \frac{i}{2}(\ln(1+i)-\ln(1-i))=-\frac{\pi}{4}$$ which means $$F(z) = \arctan z - \frac{\pi}{2}$$

In fact, the correct (principal) value of $\arctan z$, for all $z$ except the branch cut, is $$\arctan z = \frac{i}{2}(\ln(1-iz)-\ln(1+iz))$$

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  1. The expression $$F(z)=\frac{i}{2}\log(z+i)-\frac{i}{2}\log(z-i)$$ uses logs. Logs are bad news. We know that in order to use log we must either have a multivalued function, or pick a branch.

    We can pick the usual branch by cutting along the negative real axis. That means that $\log z$ is defined, but only when $\operatorname{Re} z > 0$. $+i$ and $-i$ don't affect this. So that's why the question includes the "$\operatorname{Re} z > 0$" condition.

  2. You are right that just having the same derivative does not necessarily make it equal to $\arctan$. However, it might just happen to be equal. So you will have to plug in a value of $z$ and see what happens -- $z = 1$ might be good. Of course you will have to remember the branch of $\log$ that we are using, and also consult your book or class for the branch of $\arctan$ that you are using.

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  1. Observe that $Re(z) > 0 \iff Arg(z) \in [\frac{-\pi}{2}, \frac{\pi}{2}]$: In the right half of $\mathbb C$, $Ln(z)$ is differentiable. The point is to exclude $\mathbb R_{\le 0}$. So we could still have the same result for $Arg(z) \in [\frac{-\pi}{4}, \frac{\pi}{4}]$ and $Arg(z) \in [\frac{-2\pi}{3}, \frac{2\pi}{3}]$.

  2. If you indeed got this from A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 4.17, then you'll observe they actually don't have an explicit definition for $\arctan(z)$. On Wiki the definition they give is equivalent to $$\frac{i}{2}(Ln(1-iz)-Ln(1+iz)).$$ All we have to do is check if that's equal to $F(z)$:

$$F(z) : = \frac{i}{2}Ln(z+i)-\frac{i}{2}Ln(z-i) = \frac{i}{2}(Ln(1-iz)-Ln(1+iz))$$

$$\iff Ln(z+i)-Ln(z-i) = Ln(1-iz)-Ln(1+iz)$$

$$\iff \ln|z+i|-\ln|z-i| = \ln|1-iz|-\ln|1+iz| \label{a}\tag{1}$$

and

$$Arg(z+i)-Arg(z-i) = Arg(1-iz)-Arg(1+iz) \label{b}\tag{2}$$

Now $\ref{a}$ is true but $\ref{b}$ is false.

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