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  • Can someone explain to me, step by step, how to calculate this,

$$x=(-1-i)^{15}+(-1+i)^{11}$$

enter image description here

  • Method 1 (Transform the numbers (−1−i) and (−1+i) to polar coordinates ) by Mr 5xum

\begin{align*} x&=(\sqrt{2})^{15}\exp(-\frac{45\pi}{4}i)+(\sqrt{2})^{11}\exp(\frac{33\pi}{4}i)\\ x&=128\sqrt{2}\exp(-\frac{45\pi}{4}i)+32\sqrt{2}\exp(\frac{33\pi}{4}i) \end{align*}

but i got stuck

  • Method 2

note that $(1+i)^{2}=2i ,\quad (i-1)^{2}=-2i$ then :

\begin{align*} x&=(-1)^{15}(1+i)^{15}+(i-1)^{11}\\ x&=-(1+i).(1+i)^{2.7}+(i-1)(i-1)^{2.5}\\ x&=-(1+i).(2i)^{7}+(i-1)(-2i)^{5}\\ x&=-(1+i).(-128i)+(i-1)(-32i)\\ x&=(-128+32)+(128+32)i\\ \end{align*} enter image description here

this is easy way but what about the first method can someone explain to me, in details how to use it

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    $\begingroup$ $Q^{15}=Q^{14}Q=(Q^2)^7Q$, similarly $L^{11}=(L^2)^5L$. $\endgroup$ May 8, 2014 at 13:16
  • $\begingroup$ Do you know what $e^{2\pi i}$ is? Can you see how that helps you to evaluate $e^{33\pi i/4}$? $\endgroup$ May 9, 2014 at 9:25
  • $\begingroup$ $e^{2\pi i}=1$ more we've $e^{2\pi k.i}=1 \quad \forall k \in \mathbb{Z}$ $\endgroup$
    – Educ
    May 9, 2014 at 13:35
  • $\begingroup$ Yes. So, what about $e^{33\pi i/4}$? $\endgroup$ May 11, 2014 at 12:51

1 Answer 1

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Hint: Transform the numbers $(-1-i)$ and $(-1+i)$ to polar coordinates first. Then calculate their powers.

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  • $\begingroup$ so is this the trick to answer that kind of questions ? $\endgroup$
    – Educ
    May 8, 2014 at 11:22
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    $\begingroup$ @Educ Yes! Also to solve questions like $x^5=i$ $\endgroup$
    – Bernhard
    May 8, 2014 at 11:23
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    $\begingroup$ Few steps:$-1-i=\sqrt(2) cis(-\frac{3 \pi}{4})$ $\Rightarrow (-1-i)^{15} = (\sqrt(2))^{15}cis(-\frac{45\pi}{4})$ And, $-1+i=\sqrt(2) cis(\frac{3\pi}{4})$ $\Rightarrow (-1+i)^11=(\sqrt(2))^{11}cis(\frac{33\pi}{4})$ $\endgroup$
    – nam
    May 8, 2014 at 11:25
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    $\begingroup$ @Educ Yes! In short, when you are adding two complex numbers, it is easiest to add $a+bi$ to $c+di$ to get $a+c + (b+d)i$. When you are multiplying the numbers or taking their powers, it is easier to multiply $r(\cos\alpha + i\sin\alpha) \cdot s(\cos\beta + i\sin\beta) = rs(\cos(\alpha+\beta) + i\sin(\alpha + \beta))$. $\endgroup$
    – 5xum
    May 8, 2014 at 12:21

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