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


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

  • 1
    $\begingroup$ $Q^{15}=Q^{14}Q=(Q^2)^7Q$, 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


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

  • $\begingroup$ so is this the trick to answer that kind of questions ? $\endgroup$
    – Educ
    May 8, 2014 at 11:22
  • 1
    $\begingroup$ @Educ Yes! Also to solve questions like $x^5=i$ $\endgroup$
    – Bernhard
    May 8, 2014 at 11:23
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.