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Find a branch of $\log(z^2+1)$ that is analytic at $z=0$ and takes the value of $2\pi i$ there. Also, determine a branch of $\log(z^2+2z+3)$ that is analytic at $z=-1$.

If I plug in $z=0$ and $z=-1$ to there respective functions then I get $\log(1)$ and $\log(2)$ but then what do I have to do to find a branch?

Define a branch of $(z^2-1)^{1/2}$ that is analytic in the exterior of the unit circle $|z|>1$.

If I transform this function into $e^{(1/2)Log(z^2-1)}$, why will the principal branch not work here?

Ted
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Q.matin
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2 Answers2

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For your first question: To find a branch of $\log(z^2+1)$ means to find an analytic function $f$ such that $\exp(f(z)) = z^2+1$. Here you haven't specified what the domain of $f$ should be, but presumably it should be as large as possible (and of course containing 0, which is stated in the problem).

Hint: Factor $z^2+1$ and use the standard properties of logarithms from the real numbers. Even though these properties don't hold in all cases for complex logarithms, they will help you to guess the answer, which you can then prove directly by exponentiating, as I described in the answer to your other question.

For your second question: The principal branch does not work because for some values of $z$ in $|z|>1$ (for example $z=2i$), the value of $z^2-1$ will be a negative real number, where $Log$ is not defined.

Ted
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  • Thanks Ted, but I still do not know how to do the first part. The $z$ is throwing me off. Can you give some more hints? Also, to the second question, what branch will work then if not the prinicpal? – Q.matin Jan 28 '13 at 06:35
  • What do you mean by "the $z$ is throwing me off" ? How far did you get with the hint? – Ted Jan 28 '13 at 06:57
  • What I meant by that is I know how to find a branch of , e.g log(1) or log(i), but I haven't seen an example where z is in log. And I didn't get far at all with the hint. – Q.matin Jan 28 '13 at 07:53
  • In the other question you posted, you proved that $Log(-z) + i \pi$ is a branch of $\log(z)$, by showing that $\exp(Log(-z)+i\pi) = z$. This is similar, only you're not given the function this time, you have to find it, and show that it exponentiates to $z^2+1$. – Ted Jan 28 '13 at 07:59
  • So I have to show this $\exp(Log(z^2 +1)$? – Q.matin Jan 28 '13 at 08:07
  • Also, I am still not understanding the second part. What if it were in the interior of the circle instead of the exterior: $|z|<1$? – Q.matin Jan 28 '13 at 08:12
  • $Log(z^2+1)$ is a branch of $\log(z^2+1)$, but the problem also says that the branch has to take the value $2 \pi i$ at $z=0$. Does $Log(z^2+1)$ at take the value $2 \pi i$ at $z=0$? (No it doesn't) So you have to choose a different branch. – Ted Jan 28 '13 at 08:12
  • For the second part: If it were the interior, the principal branch $Log(z^2-1)$ still does not work because e.g. at $z=0$, the value of $z^2-1$ is a negative real number. – Ted Jan 28 '13 at 08:14
  • Does then $Log(z^2+1)$ take the value of $\pi i$ at z=0? – Q.matin Jan 28 '13 at 08:18
  • I hope I am not annoying you but if you dont mind Ted, I want to ask a more specific question to this because I really want to understand. For part 2, why exactly does it has branch cuts in the y-asix? – Q.matin Jan 28 '13 at 08:26
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Let me give a simpler function so you can see what's going on. Let's consider the function $f(z)=\ln(z)$. We would like to find a branch where $\ln(z)$ is analytic at $z=1$ and has a value $2\pi i$ there. Now, you know that

$$ \ln(z)=\ln|z|+i(\theta+2k\pi) \longrightarrow (1). $$

We want to find the branch of the function that has value $2\pi i$ at

$$z=1 \implies z=e^{0i} \implies \theta = 0 . $$

Substituting $\theta =0 $ in $(1)$, $z=1$, and choosing the right branch, in our case $k=1$, we have

$$ \ln(1)= 0 + i(0+2\pi)=2\pi i. $$

I hope this example makes it clear for you.

Mhenni Benghorbal
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    Thanks a lot! It does clear some issues. You got $\theta = 0$ because $ln(1)=0$, is that correct? Also, what exactly does it mean when you said has a value of $2\pi i$, I want to understand it intuitively ? – Q.matin Jan 28 '13 at 08:58
  • @Q.matin: We get $\theta =0$ by writing the complex number $z$ in polar coordinates. In our case, the complex number is $z=1$. Remember that $z=|z|e^{i\theta}$. – Mhenni Benghorbal Jan 28 '13 at 09:04
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    Ohh okay, I see now since $z=1$ iff $z = e^{\theta i}$ , where $\theta = 0$. Thanks a lot for clearing that up for me!! Last question, what exactly does it mean when you said has a value of 2πi? Does that mean the arg must equal $2πi$? – Q.matin Jan 28 '13 at 09:20
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    @Q.matin: It is the value of $\ln(z)$ at the point $z=1$. – Mhenni Benghorbal Jan 28 '13 at 12:27
  • Well, I take it that you assumed, in your text, $-\pi<\theta<\pi$. Is it the case ? – Duchamp Gérard H. E. Dec 31 '21 at 13:25