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^{j}b^{{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 ik\right),k\in {\mathbb  {Z}}\,

i^{{p-j}}=\exp \left(i{\frac  {\pi }{2}}(p-j)+2\pi ik(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]\,


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