Originally Posted by kjchauhan

\displaystyle \begin{align*} \left( \alpha + i\,\beta\right)^{x + i\,y} &= \left( r\,e^{i\theta} \right)^{x + i\,y} \\ &= r^{x + i\,y}\,e^{i\theta\left(x + i\,y\right)} \\ &= r^x\,r^{i\,y}\,e^{-y\,\theta + i\,\theta\,x} \\ &= r^x\,r^{i\,y}\,e^{-y\,\theta}\,e^{i\,\theta\,x} \\ &= r^x\,e^{\log{\left( r^{i\,y} \right)}}\,e^{-y\,\theta}\,e^{i\,\theta\,x} \\ &= r^x\,e^{i\,y\log{r}}\,e^{-y\,\theta}\,e^{i\,\theta\,x} \\ &= r^x\,e^{-y\,\theta}\,e^{i\left(y\log{r} + \theta\,x\right)} \\ &= r^x\,e^{-y\,\theta}\left[\cos{\left(y\log{r} + \theta\,x\right)} + i\sin{\left(y\log{r} + \theta\,x\right)}\right] \end{align*}

Thank you very much for proof..

This proof is very good.And it easily understandable.

Thank you
Bizworldusa

I don't really have any others like this one. The reason is the only courses I have taught were Calculus, and this was intended as a review sheet for incoming freshman who were taking my class. I do have problem-solving stickies in the Other Topics and Advanced Applied Math forums. snapback obey snapback new era pas cher casquette tisa