Is there a systematic way to express the sum of two complex numbers of different magnitude (given in the exponential form), i.e find its magnitude and its argument expressed in terms of those of the initial numbers?
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Convert to rectangular form, add. convert back.
$$\begin{eqnarray} [r, \theta] + [s, \phi]& \to& (r \cos \theta, r \sin \theta) + (s \cos \phi, s \sin \phi) \to (r \cos \theta + s \cos \phi, r \sin \theta + s \sin \phi)\\ &\to& \bigg[\sqrt{(r \cos \theta + s \cos \phi)^2 + ( r \sin \theta + s \sin \phi)^2},\\ &&\text{atan2}(r \sin \theta + s \sin \phi, r \cos \theta + s \cos \phi)\bigg] \end{eqnarray}$$
where I'm using brackets for polar form, and parens for $x + iy$ form. "atan2" returns an arg between $-\pi$ and $\pi$, so you may want to add $2\pi$ if the returned arg is negative.
The expression under the square-root can be expanded and simplified somewhat, but I find it's probably simpler to see what's going on in this form.
fgp
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John Hughes
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Okey thx. I wanted actually to simplify this expression $$(i-1)^{11}-(i+1)^{15}$$ WA says it's equal to $-96+160i$ – aflous May 08 '14 at 12:18
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1@aflous Note that $(i+1)^2=2i$ and $(i-1)^2=-2i$. – Lutz Lehmann May 08 '14 at 12:25
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Thx. I used to know this identity but didn't notice it here :) – aflous May 08 '14 at 12:26
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Not really. One has to express the complex numbers as the sums of their real and imaginary parts, and then add componentwise, like usual. The best we can really do is make use of the triangle inequality:
$| |z_1| - |z_2| | \leq |z_1 + z_2| \leq |z_1| + |z_2|$
Alex G.
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