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In a problem that I was given it asks to show that $\text{ln}(i^3) = 3\cdot\text{ln}(i)$ where this lower case "$\text{ln}$" is defined as $\text{ln}(z) = \text{log}_e|z| + i\cdot\text{arg}(z)$, since $\text{arg}(z) = \theta+2n\pi$ there is a periodicity to the natural log function.

If I carry out this problem I would get the conclusion that $\text{ln}(i^3) = i\cdot(-\frac{\pi}{2} + 2n\pi)$ and $\text{ln}(i) = i\cdot(\frac{\pi}{2} + 2k\pi)$ (using k since it can be different than n). Multiplying the latter by 3 would yield, $3\cdot\text{ln}(i) = i\cdot(\frac{3\pi}{2} + 6k\pi)$ but since $n, k$ are integers how can we say this is equal? It seems that $(-\frac{\pi}{2} + 2n\pi)$ "hits" all the values that $(\frac{3\pi}{2} + 6k\pi)$ would but not vise versa. Is the problem wrong or is there something about the periodicity that I am not understanding in this context?

Conor_Meise
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2 Answers2

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Multiplying the later by 3 would yield, $3\cdot\text{ln}(i) = i\cdot(\frac{3\pi}{2} + 6k\pi).$

The above excerpt is accurate, but misleading.

You have determined that one of the satisfying values of $~z~$ such that $~e^z = i^3,~$ is $~z = i(\theta),~$ where $~\theta = 3\pi/2~$ which is congruent $\pmod{2\pi}~$ to $~-\pi/2.~$

Given any $~\theta \in \Bbb{R},~$ you have that $~e^{i\theta} = e^{i(\theta + 2k\pi)} ~: ~k \in \Bbb{Z}.~$

This implies that the set of satisfying values of $~z~$ is given by $~z = i\theta,~$ where $~\theta~$ refers to the set of values $~(-\pi/2 + 2k\pi) ~: ~k \in \Bbb{Z},~$ rather than the (mistaken) $~(-\pi/2 + \color{red}{6}k\pi).$

As an illustration, note that both $~z = i(-\pi/2),~$ and $~z = i(3\pi/2),~$ are satisfying values of $~z.$

Conor_Meise
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user2661923
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  • I don't know if I follow, the problem is that for ln(i) the set of values would be (-pi/2 +2kpi) and that is being multiplied by 3 – Conor_Meise Jan 31 '24 at 01:51
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    @Conor_Meise You are correct to take the original log of $~i~$ which is $~i(\pi/2),~$ and multiply it by $~3~$ to get a single satisfying value of $~z,~$ that might be expressed as (for example) $~i(3\pi/2).~$ However, once such a satisfying value of $~z~$ is found, the set of all satisfying values of $~z~$ must be expressed as $~i(\theta),~$ where $~\theta~$ is any real number that is congruent, mod $~2\pi,~$ to $~3\pi/2.~$ ...set next comment – user2661923 Jan 31 '24 at 01:55
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    @Conor_Meise Your mistake was in thinking that after a single satisfying value of $~z~$ was found, that the set would be given by $~i(\theta + 6k\pi),~$ simply because you had to originally multiply $~i\pi/2~$ by $~3~$ to get the first value $~3\pi/2.~$ You can not escape that for any $~\theta,~$ that $~\displaystyle e^{i\theta} = e^{i(\theta + 2k\pi)} ~: ~k \in \Bbb{Z}.$ – user2661923 Jan 31 '24 at 01:57
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    @Conor_Meise Another way of saying the same thing is to notice that $~\left[ ~e^{i\pi/2} ~\right]^3 = \left[ ~e^{-i\pi/6} ~\right]^3.$ So, since you are cubing a value $~w = e^{i\alpha},~$ the set of all values that will result in the same cube are given by $~e^{i(\alpha + 2k\pi/3)} ~: ~k \in \Bbb{Z}.$ – user2661923 Jan 31 '24 at 02:01
  • Okay, thank you for your help! – Conor_Meise Jan 31 '24 at 02:16
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"$\ln$" is defined as $\ln(z) = \log_e|z| + i\cdot\arg(z)$

is not really a function $\mathbb C \backslash\{0\} \to \mathbb C$ if you allow multiple real values to be possible arguments.

Instead split $\arg(z)$ to make it a function and try one of:

  • (A) "$\ln$" is defined as $\ln(z) = \log_e|z| + i\cdot\arg(z)$ where $\arg(z)\in \left(-\pi, \pi\right]$
  • (B) "$\ln$" is defined as $\ln(z) = \log_e|z| + i\cdot\arg(z)$ where $\arg(z)\in \left(0, 2\pi\right]$

With (A) you get $\ln(i^3)=\ln(-i)=i\cdot\frac{-\pi}2=-\ln(i)$

while with (B) you get $\ln(i^3)=\ln(-i)=i\cdot\frac{3\pi}2=3\ln(i)$

and you need to be using definition (B) for your result.

Note that both A and B lead to $\ln(i^5)=\ln(i)$ rather than $\ln(i^5)=5\ln(i)$.

Henry
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