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so i have this complex number $\dfrac1{(1+i)^{100}}$ and i need to use De Moivre's formula formula to power. So how do you calculate it ? For what i know it should go like this :
$\dfrac1{(1+i)^{100}} = (1+i)^{-100}$
$a = 1 , b =1 $
$n = -100$
$r=2^{1/2}$
$\tan(\frac11) = 45^\circ$
so we get $z = 2^{1/2} ( \cos(45^\circ) + i\sin(45^\circ))$
$(2^{1/2})^{-100} ( \cos(-4500^\circ) + i\sin(-4500^\circ))$ =
$2^{-50} ( (-1) + i\cdot0) = -\dfrac{1}{2^{50}}$ ???
so i fixed the mistake of $45*100$
$cos(-4500) = -1 , sin(-4500) = 0 $ ?

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  • $\begingroup$ It is $-4500$ degrees. $\endgroup$ – André Nicolas Aug 28 '13 at 23:51
  • $\begingroup$ Other than $-45^\circ\times100=-4500^\circ$, yes. $\endgroup$ – robjohn Aug 28 '13 at 23:52
  • $\begingroup$ If we assume the incorrect $-450$, the calculation is not right, since the sine of $-450$ degrees is $-1$. $\endgroup$ – André Nicolas Aug 28 '13 at 23:54
  • $\begingroup$ Now it is correct, $-\frac{1}{2^{50}}$. $\endgroup$ – André Nicolas Aug 29 '13 at 0:58
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$\dfrac1{{(1+i)}^{100}} =(1+i)^{-100}=(\sqrt2(\cos45^\circ+i\sin45^\circ))^{-100}$.

Now use De Moivre's formula.

In your steps, you have accidentally written $(\cos-450^\circ+i\sin-450^\circ)$ instead of $(\cos-4500^\circ+i\sin-4500^\circ)$.

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