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I'm studying complex numbers and I wanted to know if this solution is correct. The problem is to calculate $(1+i)^{11}$, here's my attempt:

I can express $(1+i)^{11}$ using the argument notation: $z=\rho(\cos \phi +i\sin \phi)$

$$z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$$

This is very helpful because I can use De Moivre's formula to calculate $z^{11}$ $$z^{11}=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{11}=\left(\cos \frac{11\pi}{2}+i\sin\frac{11\pi}{2}\right)=0-i=-i$$

Is this correct? Is there a better way to solve it?

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    $\begingroup$ Do you know the complex exponential notation? $\endgroup$
    – Bernard
    Sep 10, 2016 at 14:59
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    $\begingroup$ $\cos\frac{\pi}{2}$ is $0$, not $1$. $\endgroup$
    – TonyK
    Sep 10, 2016 at 15:39

6 Answers 6

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It is wrong when you write $z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$ (actually this number is $z=i$), indeed $z=\sqrt2\left(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$.

Finally you get : $$z^{11}=\sqrt2^{11}\left(\cos \frac{11\pi}{4}+i\sin\frac{11\pi}{4}\right)$$ $$=2^5\sqrt2\left(\cos \frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$$ $$=2^5\sqrt2\left(\frac{-\sqrt2}{2}+i\frac{\sqrt2}{2}\right)$$ $$=2^5(i-1)=-32+32i$$

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  • $\begingroup$ Is this correct? $$(\sqrt{2})^{11}\left(\cos \frac{11\pi}{4}+i\sin\frac{11\pi}{4}\right)=(\sqrt{2})^{11}\left(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=(\sqrt{2})^{10}\cdot(-1+i)$$ $\endgroup$
    – PunkZebra
    Sep 10, 2016 at 14:57
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    $\begingroup$ You can simplify: $\sqrt{2}^{11}=2^5$… $\endgroup$
    – Bernard
    Sep 10, 2016 at 14:59
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    $\begingroup$ @Bernard: You mean $\sqrt 2^{10}=2^5$. $\endgroup$
    – TonyK
    Sep 10, 2016 at 15:37
  • $\begingroup$ @Tony/ Not quite. Actually, I meant the final $\sqrt 2^{10}=2^5$. Thanks for pointing the typo. $\endgroup$
    – Bernard
    Sep 10, 2016 at 16:02
  • $\begingroup$ @peterix ,you're right! $\endgroup$ Sep 10, 2016 at 16:52
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It should be $$z=\sqrt{2}e^{i\pi/4}$$ So $$z^{11}=2^{11/2}\left(\cos\frac{11\pi}{4}+i\sin\frac{11\pi}{4}\right)$$

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  • $\begingroup$ No, it's the OP's $z$. $\endgroup$
    – velut luna
    Sep 10, 2016 at 14:38
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We can do that without trigonometry. Just notice that:

$$(1+i)^{11} = [(1+i)^2]^5 \cdot (1+i)$$

So:

$$ (i+1) \cdot (i^2 + 2i + 1)^5$$

From that:

$$(1+i)\cdot (2i)^5$$

But $i^5 = (- 1)\cdot (-1)\cdot i$

So:

$$[(2^5)\cdot i] \cdot (i+1) = - 32+32i$$

That equals:

$$32(i-1)$$

I find that way easier to solve, but is up to you. Good luck!

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You might note that $(1+i)^2 = 2i$, so that $(1+i)^{10} = 2^{5}i^{5} =2^{5}i$. Then $(1+i)^{11} = -2^{5}i(1+i)= 32 - 32i.$

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  • $\begingroup$ Since the downvoter doesn't seem to have explained: it should be $(1 + i)^{10} = (2i)^5 = 2^5 i$. $\endgroup$ Sep 10, 2016 at 15:35
  • $\begingroup$ Why they vote down this answer? This is nice $\endgroup$ Sep 10, 2016 at 16:58
  • $\begingroup$ Internet hazing. Not very professional, but they think it's their "culture" and that it justifies bullying. $\endgroup$
    – B. Goddard
    Sep 10, 2016 at 17:02
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    $\begingroup$ I don't think it's got anything to do with hazing. Initially, the answer was incorrect; that's a perfectly valid reason to downvote. They probably should have commented though, because otherwise the odds of them noticing the correction and retracting their vote is pretty slim... $\endgroup$ Sep 10, 2016 at 17:28
  • $\begingroup$ Well, sometimes a typo is just mentioned. Other times it gets nuked. There's no consistency. Well, it there's consistency in that the nuking is always anonymous. There's certainly an "inner circle" here and they don't make things better. $\endgroup$
    – B. Goddard
    Sep 10, 2016 at 18:14
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There's another approach:

\begin{align} (1+i)^{11}&=\sum_{k=0}^{11}\binom{11}{k}i^k \\\\&=\sum_{\substack{0\le k\le 11 \\k\equiv 0\!\pmod 4}}\binom{11}{k} + \sum_{\substack{0\le k\le 11 \\k\equiv 1\!\pmod 4}}\binom{11}{k}\cdot i + \sum_{\substack{0\le k\le 11 \\k\equiv 2\!\pmod 4}}\binom{11}{k} \cdot (-1)+ \sum_{\substack{0\le k\le 11 \\k\equiv 3\!\pmod 4}}\binom{11}{k}\cdot (-i)&\scriptsize\text{by the binomial theorem, and computing each }i^k \\\\&=\sum_{k=0,4,8}\binom{11}{k}+i\sum_{k=1,5,9}\binom{11}{k}-\sum_{k=2,6,10}\binom{11}{k}-i\sum_{k=3,7,11}\binom{11}{k} \\\\&=\sum_{k=0,4,3}\binom{11}{k}+i\sum_{k=1,5,2}\binom{11}{k}-\sum_{k=2,5,1}\binom{11}{k}-i\sum_{k=3,4,0}\binom{11}{k}&\scriptsize\text{since }\binom{11}{k}=\binom{11}{11-k} \\\\&=\left(\sum_{k=0,3,4}\binom{11}{k}-\sum_{k=1,2,5}\binom{11}{k}\right)(1-i) \\\\&=\left(1 + \frac{11\cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}+ \frac{11\cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} -11 -\frac{11\cdot 10}{2 \cdot 1}-\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\right)(1-i) \\\\&=\left((1 + 165+ 330)-(11+55+462) \right)(1-i) \\\\&=-32(1-i) \\\\&=-32+32i \end{align}

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it is too simple, $z=(i+1)^11$

$= [(i+1)^2]^5 (i+1);$

$= [2i]^5 x (i+1)$

$= 32i (i+1)$

$= -32+32i$

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