9/10 people get this wrong

1. Aug 6, 2010

elfboy

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

Last edited: Aug 6, 2010
2. Aug 6, 2010

Mentallic

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

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

3. Aug 6, 2010

gomunkul51

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).

Last edited: Aug 6, 2010
4. Aug 6, 2010

hgfalling

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

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

5. Aug 6, 2010

hgfalling

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

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.

6. Aug 6, 2010

jackmell

You made no mention of principal value so the real part is actually multi-valued:

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

7. Aug 6, 2010

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8. Aug 6, 2010

elfboy

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

9. Aug 6, 2010

DaveC426913

I am in the top 10 percentile of this teaser.

I definitely did not get it wrong.

10. Aug 6, 2010

Dick

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?

11. Aug 7, 2010

Mentallic

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 :/

Last edited: Aug 7, 2010
12. Aug 7, 2010

Staff: 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)?

13. Aug 7, 2010

Mentallic

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