CV8

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Evaluate $\mathrm{Re}\left[(a+bi)^p\right]\,$ .

with complex variables

Write $(a+bi)^p = e^{\tan^{-1}p\frac{b}{a}}\,$ .

Now $\mathrm{Re}\left[e^{\tan^{-1}p\frac{b}{a}}\right] = \cos(p\tan^{-1}\frac{b}{a})\,$ .

with regular calculus

The binomial theorem and Pochhammer symbols give another way to write the quantity

$(a+b)^p = \sum_{j=0}^\infty {p \choose j}a^j b^{p-j} = \sum_{j=0}^\infty\frac{(-1)^j}{j!}(-p)_j a^j b^{p-j}\,$

So,

$\mathrm{Re}\left[(a+bi)^p\right] = \mathrm{Re}\sum_{j=0}^\infty \frac{(-1)^j}{j!}(-p)_j a^j(bi)^{p-j}\,$

$=\sum_{j=0}^\infty \frac{(-1)^j}{j!}(-p)_j a^jb^{p-j}\mathrm{Re}\left[i^{p-j}\right]\,$

$=b^p\sum_{j=0}^\infty \frac{\left(-\frac{a}{b}\right)^j}{j!}(-p)_j \mathrm{Re}\left[i^{p-j}\right]\,$

Write $i = \exp\left(i\frac{\pi}{2} + 2\pi i k\right), k\isin\mathbb{Z}\,$

$i^{p-j} = \exp\left(i\frac{\pi}{2}(p-j) + 2\pi i k(p-j)\right)\,$

$\mathrm{Re}\left[i^{p-j}\right]=\cos\left[\frac{\pi}{2}(p-j)+2\pi k (p-j)\right]\,$

Finally,

$\mathrm{Re}\left[(a+bi)^p\right]=b^p\sum_{j=0}^\infty \frac{\left(-\frac{a}{b}\right)^j}{j!}(-p)_j \cos\left[\frac{\pi}{2}(p-j)+2\pi k (p-j)\right]\,$

Main Page : complex Variables

CV8

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with complex variables

with regular calculus

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