Complex logarithm

Algebra Level 3

Which of the options is a value for $\large \log_{i} e?$

-\dfrac{i}{2\pi}

-\dfrac{2i}{\pi}

None of these choices

-\dfrac{i}{\pi}

2 solutions

Surya Prakash
Nov 29, 2015

$i = \cos \left(\pi / 2 \right) + i \sin \left(\pi/ 2 \right)$

Using euler's representation, we can write $i$ as $i = e^{i \pi /2}$ .

So, $\ln i =\dfrac{ i \pi}{2}$ .

So,

$\log _{i} e = \dfrac{\ln e}{\ln i} = \dfrac{1}{i \dfrac{\pi}{2}} = \dfrac{2}{i \pi} = \dfrac{-2i}{\pi}$

Moderator note:

As always, with complex exponentiation and logarithm, the answer that you get isn't a single value. Instead, it could be multi-valued. Refer to Mark's comment to understand what all the possible values are.

I have to disagree with the Challenge Master (if only in detail). The logarithm is multivalued, and there are infinitely many possible answers. On the other hand, $\tfrac{2}{\pi}i$ is not one of them. We are looking for a complex number $z$ such that $i^z = e$ . We could have $\left(e^{(2n+\frac12)\pi i}\right)^z \; = \; e \; = \; e^1$ for any $n \in \mathbb{Z}$ , depending on what argument we want to assign to $i$ . This means that we could, possibly, have $(2n+\tfrac12)\pi i z \; = \; 1 + 2m\pi i$ for any $m,n \in \mathbb{Z}$ . This gives us the range of possible answers for $z$ : $\tfrac{4m}{4n+1} - \tfrac{2}{(4n+1)\pi}i \qquad \qquad m,n \in \mathbb{Z} \;.$ This collection does not include $\tfrac{2}{\pi}i$ .

While this means that there is only one of the proposed answers to this question that can be correct, it is misleading to imply that $\log_ie$ can be evaluated . Better ask which of the numbers listed could be a value for the logarithm.

Logarithms of complex numbers have to be treated very carefully; they can only be uniquely defined once a clear definition of argument has been understood. Trying to define logarithms to a complex base is doubly complicated, since we have to determine the argument of the base as well as that of the number whose logarithm we are taking!

Mark Hennings - 5 years, 6 months ago

Yes you are right. I did my calculation too quickly and simply replaced $\pi$ with $- \pi$ . Sorry about that. Let me edit it accordingly.

Agreed about stressing that we have a multi-valued function in the problem statement.

Calvin Lin Staff - 5 years, 6 months ago

For $z\in \Bbb C$ $\ln z=\ln |z|+Arg(z)$ but this gives us $\frac{2}{\pi}$ as a solution so?

Miliyon Tilahun - 5 years, 6 months ago

There are multiple values for complex exponential and complex logarithms.

Check through your work, $\frac{2}{ \pi}$ is not one of the values (I made the same mistake too!). You can see Mark's comment to understand how to determine all of these values.

Calvin Lin Staff - 5 years, 6 months ago

Lu Chee Ket
Dec 6, 2015

$\log_ie = \displaystyle \frac{\ln e}{\ln i} = \displaystyle \frac{1}{\ln e^{\frac{\pi}{2}i}} = \displaystyle \frac{1}{\frac{\pi}{2}i} = \frac{-2 i}{\pi}$

Complex logarithm

2 solutions

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