Finding Complex root of z^{3/4} types

Question

Can't get more than one solution whilst going over this textbook question:

$$\ z^{3/4}= \sqrt{6}\ +\sqrt{2}i $$ $$\text{Answers in the form } re^{i\theta}; r>0,\text{ and } -\pi < \theta \leqslant \pi$$

Converted the RHS ($\ z^\frac{3}{4}=\sqrt{8}e^{\frac{\pi}{6}i}$), then applied de Moivre's Theorem and introduced $2k\pi$. $(z=[\sqrt{8}e^{(\frac{\pi}{6} + 2k\pi)i}]^\frac{4}{3}$), $k\in \mathbb{z}$. So we get $$\\z=4e^{\frac{4}{3}i(\frac{\pi}{6} + 2k\pi)}$$ $$ z=4e^{i(\frac{2\pi}{9} + \frac{24k\pi}{9})}$$ On inspecting this, I see the only possible value of k is $0$ as others give $\theta$ values outside the range. My answer being$\ z=4e^{\frac{2\pi}{9}i}$.Where have I gone wrong? I'm sure there's more than one solution, the textbook indicates so.

PostScript:$\ $ So, in the solution that's provided in the text book, de Moivre's Theorem is applied twice: first using $[\sqrt{8}e^{i(\frac{\pi}{6})}]^4=64e^{i(\frac{2\pi}{3})}$ and then $[64e^{i(\frac{2\pi}{3}+2k\pi)}]^\frac{1}{3}=4e^{i(\frac{2\pi + 6k\pi}{9})}$ It's on the second application that they introduced the $2k\pi$. Using their method, we reach 3 solutions, one of which is the one I got.

Why can't de Moivre's theorem be applied in one step with the whole fraction?

Answer 1

It can, and your solution is correct... up to the point where you concluded that the "others give $\theta$ values outside the range". The range doesn't matter: the solutions are actually correct for every $k\in\mathbb Z$, except that it turns out that some of them are identical, so we try to remove the duplicates and keep just the non-identical ones.

To demonstrate: in the book, the solutions are $4e^{i\frac{2\pi}{9}}$, $4e^{i\frac{8\pi}{9}}$ and $4e^{i\frac{14\pi}{9}}$. Let's try to see that you got the same three solutions using your formula, by varying $k$:

• $k=0$: $4e^{i\frac{2\pi}{9}}$
• $k=1$: $4e^{i\frac{26\pi}{9}}=4e^{i\left[\frac{18\pi}{9}+\frac{8\pi}{9}\right]}=4e^{i\left[2\pi+\frac{8\pi}{9}\right]}=4e^{i\frac{8\pi}{9}}$
• $k=2$: $4e^{i\frac{50\pi}{9}}=4e^{i\left[\frac{36\pi}{9}+\frac{14\pi}{9}\right]}=4e^{i\left[4\pi+\frac{14\pi}{9}\right]}=4e^{i\frac{14\pi}{9}}$

... the same as the book solution.

Then, $k=3$ would give you $4e^{i\frac{74\pi}{9}}=4e^{i\left[\frac{72\pi}{9}+\frac{2\pi}{9}\right]}=4e^{i\left[8\pi+\frac{2\pi}{9}\right]}=4e^{i\frac{2\pi}{9}}$, i.e. the same solution as for $k=0$, and similarly, the other solutions will appear again when $k$ takes subsequent values ($4,5,6$ etc.)

Answer 2

In my opinion, it's simpler to solve congruences:

$$\bigl(\mathrm e^{i\theta}\bigr)^{\tfrac 34 }=\mathrm e^{\tfrac{i\pi}6}\iff \frac{3\theta}4\equiv \frac\pi6\mod 2\pi\iff9\theta\equiv 2\pi\mod 24\pi\iff\theta\equiv \frac{2\pi}9\mod \frac {8\pi}3,$$ so that $$\theta\in\Bigl\{\frac{2\pi}9,~ \frac{26\pi}9,\frac{50\pi}9\Bigr\}. $$