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How would you state $\sum_{n=1}^{135} (1-i)^n$ in standard form.

in polar form $(1-i)$ can be expressed as $\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})$ and also through De Moivre's Thereom this can be expressed as $\sqrt2(\cos(\frac{3n\pi}{4})+i\sin(\frac{3n\pi}{4})$ But from here im not sure how i would apply the sums to 135

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One has for $q\neq 1$

$$\sum_{n=1}^Nq^n=q\cdot{1-q^N\over 1-q}$$

Replace $q$ with $1-i =\sqrt{2}e^{i\pi\over 4}$. One gets

$$\sum_{n=1}^N(1-i)^N=(1-i){1-(1-i)^N\over i}$$

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  • $\begingroup$ is this a theorem? $\endgroup$ – Skrrrrrtttt Nov 16 '17 at 0:58
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    $\begingroup$ It is just the sum of the first terms of a geometric series $\endgroup$ – marwalix Nov 16 '17 at 1:04
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The formula for the sum of a finite geometric series gives $$S:=\sum_{k=1}^{135}(1-i)^k=(1+i)\bigl((1-i)^{135}-1\bigr)\ .$$ Now $(1-i)=2^{1/2}\>e^{-i\pi/4}$ and therefore $$(1-i)^{135}=2^{135/2}\>e^{-135i\pi/4} =2^{67}\>e^{-34i\pi}\cdot 2^{1/2}e^{i\pi/4}=2^{67}\>(1+i)\ .$$ It follows that $$S=(1+i)\bigl(2^{67}(1+i)-1\bigr)=-1+\bigl(2^{68}-1\bigr)i\ .$$

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$\sum_{n=0}^{N} r^n = \frac{r^N-1}{r-1} $

so

$\sum_{n=1}^{N} r^n = \frac{r^N-1}{r-1}-r^0 = \frac{r^N-r}{r-1}$

so

$\sum_{n=1}^{135} (1-i)^n = \frac{(1-i)^{135}-(1-i)}{(1-i)-1} = i((1-i)^{135}-1+i) = (1+i)((1-i)^{134}-1)= (1+i)(2^{67}e^{i\frac{3}{4}\pi*134}-1)$

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