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The professor gave us a function $f(z) = \ln r + i \theta$ (this is for a complex analysis class). He doesn't like answering students' questions and there's no assigned textbook so I don't know where to look up such a function. How can I tell where the parentheses are supposed to go for this equation?

mr eyeglasses
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    "Wise people will stick in enough parentheses to make it impossible for anyone to mistake the meaning" - Gerry Myerson http://math.stackexchange.com/questions/33215/what-is-48%C3%B7293 – JMoravitz Mar 03 '15 at 23:18
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    It looks like it is $f(z)=\ln(r)+i\theta$. Because otherwise the lack of parenthesis is inexcusable. – Asinomás Mar 03 '15 at 23:18
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    The standard functional notation demands the use of parenthesis. – Git Gud Mar 03 '15 at 23:20
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    @ʙᴀᴅᴀᴛᴍᴀᴛʜ: Unless you have some greater context to draw from, $\ln(r)+i\theta$ is probably correct. – Regret Mar 03 '15 at 23:28
  • If one has a really sick mind, can $\ln(r)+i\theta$ be misunderstood as well, as $\ln((r)+i\theta)$? In that case maybe it is safer to say $(\ln r)+i\theta$. – Jeppe Stig Nielsen Mar 04 '15 at 09:16

3 Answers3

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Dropping the parentheses when using functional notation is done often with the trigonometric functions and logarithms, à la $\sin x$ or $\log x$. This can (and does) lead to confusion, as we can see.

Usually terms which are multiplied together are all part of the function argument, while terms added or subtracted are not:

$$\log a + b =\log(a)+b\\ \log a - b =\log(a)-b\\ \log ab + c = \log\left(ab\right)+c\\ \log\frac ab + c = \log\left(\frac ab\right)+c\\ $$

On the other hand:

$$a\log b = a\log(b)\\ \frac{\log a}b = \frac{\log(a)}b$$

By these conventions, your equation means $f(z) = \ln(r) + i \theta$.

Note that there are some exceptions. For example, $\sin x\cos x$ probably means $\sin(x)\cos(x)$ rather than $\sin(x\cos(x))$, while $\sin\omega t$ probably means $\sin(\omega t)$ rather than $\sin(\omega)t$. For reasons such as this, it is good to always use enough parentheses to make the expression unequivocal.

Regret
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    How is $\sin \omega t$ an exception? The argument is a product. – wchargin Mar 04 '15 at 04:40
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    @WChargin It's used to contrast with $\sin x \cos x$. Read "for example...while" as "for example...as opposed to" or "for example...which contrasts with" instead of "for example...and". – Caleb Stanford Mar 04 '15 at 07:08
  • @WChargin Physics has slightly different conventions than math for this. Usually, if parentheses are omitted in physics, you parenthesize all terms up until the $t$ term. –  Mar 04 '15 at 08:25
  • It's worth adding that sometimes this functional notation can go as far as $$\arctan\log\sin x.$$ I think the meaning it's clear, but maybe not for someone who never saw it. – rubik Mar 04 '15 at 10:38
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In addition to syntactic answers already discussed, you can think a little about which makes the most sense in context. For a complex analysis class, $f(z) = \ln(r) + i\theta$ is a common function to encounter. If $z$ is a complex number with modulus $r$ and principal argument $\theta$, then $f(z)$ is the principal branch of the complex logarithm, a function which is used all the time. On the other hand, $f(z) = \ln(r + i\theta)$ doesn't seem to be a particularly useful function.

Nate Eldredge
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With an expression like this, we have three possibilities:

Either

\begin{align*} f(x) &= \ln(r) + i\theta,\\ f(x) &= \ln(r + i)\theta, \text{ or}\\ f(x) &= \ln(r + i\theta). \end{align*}

We can assume the professor means $\ln(r) + i\theta$, since it requires the simplest expression in parenthesis. Generally, if the expression in parenthesis is at all 'complicated' (requires any binary operations like addition, subtraction, etc), it belongs in parenthesis. If no such operations are required, parenthesis can be safely omitted and we agree to default to the simplest argument possible for the function.

That's not to say I agree with saying something is 'obvious', but I'm merely providing hints for the future!

pjs36
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    Since the OP has stated that the question arose in a Complex Analysis course, it could be argued that the "obvious" argument for the ln function is the simple complex number. – DJohnM Mar 04 '15 at 00:06
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    @User58220: It could also be argued that it's "obviously" the complex logarithm function $\log z = \ln(|z|)+i \arg(z)$... – Hans Lundmark Mar 04 '15 at 08:28