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Express the complex number $(-1-i)^{48}(-3+i\sqrt{3})^{36}$ in the form a + bi?

I got $ (2^{24})(12^{18}) + 0i $?

Did I get it right?

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Is it $-3+\sqrt 3 i$ out there. The first term is right. – user21436 Feb 14 '12 at 19:57
In which case it is right. – user21436 Feb 14 '12 at 19:58
up vote 6 down vote accepted

You're right in both the evaluations. Good work!

Alright. Let me write a short answer:

  • Note that $(-1-i)^{48}=(1+i)^{48}$ since $48$ is even. Also, $(1+i)^2=1+2i-1=2i$ and $i^{4n}=1$ for $n \in \mathbb N$. Using these, we have that, $$\begin{align}(1+i)^{48}&=(2i)^{24}\\&=2^{24}i^{24}\\&=2^{24}\end{align}$$

  • For the second term, to suit your answer, the modified question should be, $(-3+\sqrt 3 i)^{36}$ as I suggested in the comment.

Observe that $(-3+ \sqrt 3 i)^6=-1728$ of which your result shall be a trivial consequence.

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And you get the same answer for the second term as I got? – Jim_CS Feb 14 '12 at 20:20
@JohnMcDonald Yes. – user21436 Feb 14 '12 at 20:47
@JohnMcDonald Do you need other clarifications? I have improved it to my best ability. I'll be happy to hear about where I can help. – user21436 Feb 14 '12 at 20:56

Neither of the answers posted here cites a fact that one should be aware of if one thinks about this kind of problem.

Look at $-1-i$. The angle between the positive real axis and the ray from $0$ through that number is $225^\circ$ or $-135^\circ$ (it doesn't matter which one you pick). In other words, it's half-way between the real and imaginary axes, and in a certain quadrant. When you raise that number to the $48$th power, you just add that angle $48$ times, getting $48\cdot 225^\circ = \text{exactly }30\text{ full circles with no remainder}$. So you get something pointing in the positive real direction. Then notice that $|-1-i|=\sqrt{2}$. And $\sqrt{2}^{48} = 2^{24}$.

For $-3+i\sqrt{3}$, the angle is $150^\circ$ and the absolute value is $\sqrt{3^2+\sqrt{3}^2} = \sqrt{12} = 2\sqrt{3}$.

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Note: $\:x\: =\: -1\:-\:i\: \Rightarrow\:x^4 = (2i)^2 =\: -4.\ $ Recall $\:\zeta^3 = 1\:$ for $\:\zeta = (-1 - \sqrt{-3})/2\:$

thus $\:y\: =\: \sqrt{-3} - 3\ \Rightarrow\: y^6 \: =\: \left(\sqrt{-3}\: (1+\sqrt{-3})\right)^6\ =\ (\sqrt{-3}\:(-2\:\zeta))^6 \:=\: - 2^6\cdot 3^3$

Therefore $\:x^{48} y^{36} = (x^8 y^6)^6 = (2^4\: (-2^6\cdot 3^3))^6 =\: 2^{60} 3^{18}$

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Your fighting the standard TeX spacing rules is quite characteristic! – Mariano Suárez-Alvarez Feb 15 '12 at 1:46
@Mariano Do you think the writings of Bourbaki were characteristic of any particular member? – Math Gems Feb 15 '12 at 1:49

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