Question 300834
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To calculate *[tex \Large (a\ +\ bi)^{c\ +\ di}]:


Let *[tex \Large \varphi\ =\ \arg(a\ +\ bi)\ =\ \pm\arctan\frac{b}{a}] taking the sign appropriately so that *[tex \Large a\ +\ bi\ =\ r(\cos\varphi+i\sin\varphi)]


Let *[tex \Large \theta\ =\  c\ \cdot\ \varphi\ +\ \frac{1}{2}d\ \cdot\ \ln\left(a^2\,+\,b^2\right)]


Then *[tex \Large (a\ +\ bi)^{c\ +\ di} =\ \left(a^2\,+\,b^2\right)^{c/2}\,\cdot\,e^{-d\varphi}\,\cdot\,\left(\cos\theta\ +\ i\sin\theta\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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