# 9/10 people get this wrong

evaluate the real part: (-1-2i)^(1/3)

and

show why (1-2i)^(1/3) is not equal to (-1)^(1/3)*(-1+2i)^(1/3)

you will be pulling your hair out

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Mentallic
Homework Helper
You know, I learnt something today. My hair doesn't come out easily!

14 pages of working, and I hit a dead end :grumpy:

A. The real part is about 0.2 ? :)

B. Because multiplying by (-1)^(1/3) with turn the second complex number by 45 degrees and will not give you (different quarters) (1-2i)^(1/3).

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For part a...

Well, converting this complex number to polar coordinates, we get

$$r = \sqrt{5}$$

$$\theta = \arctan 2 - \pi$$

and for the cube root we have

$$r = \sqrt[6]{5}$$

$$\theta = \frac{\arctan 2 - \pi}{3}$$

so the real part is

$$\Re z = r \cos \theta = \sqrt[6]{5} \cos \frac{\arctan 2 - \pi}{3} \approx 1.018$$

$$(1 - 2i)^\frac{1}{3} = \sqrt[6]{5} \left ( \cos \frac{-\arctan{2}}{3} + i \sin \frac{-\arctan{2}}{3}\right)$$

by the same logic as above.

$$(-1)^\frac{1}{3} (-1 + 2i)^\frac{1}{3} = \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(\sqrt[6]{5} \left ( \cos \frac{\arctan{2}}{3} + i \sin \frac{\arctan{2}}{3}\right) \approx -0.2013 + 1.292 i$$

so those things aren't equal. It's not clear why they would be, since $a^c b^c = (ab)^c$ isn't an identity for the complex numbers.

evaluate the real part: (-1-2i)^(1/3)
You made no mention of principal value so the real part is actually multi-valued:

$$(-1-2i)^{1/3}=5^{1/6}e^{i/3(-\pi+\arctan(2)+2k\pi)},\quad k=0,1,2$$

yup For part one, it's the minus pi part after the arctangent which is the part that tripped me up.

DaveC426913
Gold Member
I am in the top 10 percentile of this teaser.

I definitely did not get it wrong.

Dick
Homework Helper
I am in the top 10 percentile of this teaser.

I definitely did not get it wrong.
Yet more proof, as if any were needed, that you ARE Mr. Smartypants! The problem could be a little less trap inducing by pointing out, as jackmell did, that "^(1/3)" has to be defined. Otherwise, it's multivalued. I am curious what Mentallic spent 14 pages on. Trying to find a solution in radicals?

Mentallic
Homework Helper
I am curious what Mentallic spent 14 pages on. Trying to find a solution in radicals?
Converting into a+ib form.

More specifically, I assumed $$cos(x), sin(x)$$ where $$x=\frac{tan^{-1}2-\pi}{3}$$ wasn't a sufficient enough answer just as it wouldn't be sufficient to leave an answer as $$sin(cos^{-1}(1/2))$$ since it should be simplified.

This led me down a long and treacherous road... I thought I was going to finally get the answer in the end but it didn't seem like it was going to simplify the way I hoped. I do have an idea of how to solve it, but what I'm thinking of doing next will be the death of me. It'll require at least another 10 pages of working and simplifying :/

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Borek
Mentor
Do I classify as 1/10 or 9/10 if I have not even attempted to solve (other than entering the expression into wolfram alpha)?

Mentallic
Homework Helper
Obviously the 1/10 else if you were in the 9/10 and then checked it, you would change your answer