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I have to find the solutions (in $\mathbb{C}$, of course) of the equation $$e^{z/3} = (1 + i)^4$$

I have followed two approaches that give me different results.
I wanted to ask for your help to find the mistake I am making.

Approach 1.
I raised both sides of the equation to the power of 3, thus obtaining $$e^z = (1 + i)^{12}.$$

Simplifying and applying the logarithm to both sides \begin{align} e^z &= -2^6,\\ z &= \log(-2^6),\\ z &= 6\log|-2| + i(\arg(-2^6) + 2 k \pi),\;k \in \mathbb{Z},\\ z &= 6\log(2) + i(2 k + 1)\pi,\;k \in \mathbb{Z}.\\ \end{align}

this approach would give me solutions (showing those for $k={-2, -1, 0, 1, 2}$) $$6\ln(2) + \{..., -3 i \pi, -i \pi, i \pi, 3 i \pi, 5 i \pi,...\}$$

Approach 2.
I applied directly the logarithm to the equation without raising to the cube first \begin{align} e^{z/3} &= (1 + i)^4,\\ e^{z/3} &= -4,\\ \log(e^{z/3}) &= \log(-4),\\ \frac{1}{3} z &= \log|4| + i (2 k + 1),\; k\in \mathbb{Z},\\ z &= 3\log|4| + 3 i (2 k + 1),\; k\in \mathbb{Z},\\ \end{align}

this approach would give me solutions (showing those for $k={-2, -1, 0, 1, 2}$) $$6\ln(2) + \{..., -9 i \pi, -3 i \pi, 3 i \pi, 9 i \pi, 15 i \pi,...\}$$

As you can see the solutions of approach 2 have imaginary parts that are odd multiples of $3 i \pi$, while those of approach 1 give imaginary parts that are odd multiples of $i \pi$.

Checking with Wolfram alpha, only the solutions of approach 2 verify the initial equation. The solutions of approach 1 that do not have imaginaries parts odd multiples of $3 i \pi$ (e.g. $6\ln(2) - i \pi, 6\ln(2) + i \pi, 6\ln(2) + 5 i \pi$) do not verify it.

Now, the only difference between the two approaches is that, in approach 1, I raised both sides of the equation to the third power before applying the logarithm, so I think that is the mistake I committed. But it should be a legitimate operation, right? On the right side we have a complex number and on the left one the output of a periodic function. What have I done wrong?

cosimoNigro
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    whenever you solve a problem by raising the power on both sides by n, you would introduce excess solutions, so that you have to divide your obtained solution numbers by n. This is already obvious from real valued case of the square, where $y^2=x^2$ has solutions $y=\pm x$ as opposed to $y=x$ alone – naturallyInconsistent Apr 29 '23 at 16:45
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    Elaborating on naturallyInconsistent's comment, see this answer. – Karl Apr 29 '23 at 16:53
  • Thanks for the clarifications, as Another User points out, in $\mathbb{R}$, one can raise both sides of an equation to odd powers, this is why I thought it was fine also in this case. – cosimoNigro Apr 29 '23 at 16:55

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Your first approach assumes that $z=w\iff z^3=w^3$. This is true in $\mathbb{R}$, but not in $\mathbb{C}$. For instance $2\neq-1+\sqrt3\,i$, but $2^3=\left(-1+\sqrt3\,i\right)^3$.

Another User
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