Find real and imaginary parts

**szak1592** · March 20th 2013, 02:45 AM

e^{z^3}

and

e^1/z

where z=x+iy= r(cos theta + i sin theta)

**Prove It** · March 20th 2013, 02:56 AM

Originally Posted by szak1592

e^{z^3}

and

e^1/z

where z=x+iy= r(cos theta + i sin theta)

$\displaystyle \begin{align*} e^{z^3} &= e^{\left( r\,e^{i\theta} \right) ^3} \\ &= e^{ r^3\,e^{ 3i \theta } } \\ &= e^{ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \left[ \cos{(3\theta)} + i\sin{(3\theta)} \right] } \\ &= e^{\left( x^2 + y^2 \right) ^{\frac{3}{2}} \cos{(3\theta)} + i \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} } \\ &= e^{\left( x^2 + y^2 \right) ^{\frac{3}{2}}\cos{(3\theta)}}\, e^{ i \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} } \\ &= e^{ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \cos{(3\theta)} } \left\{ \cos{ \left[ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} \right] } + i\sin{ \left[ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} \right] } \right\} \end{align*}$

$\displaystyle = e^{ \left( x^2 + y^2 \right) ^{\frac{3}{2}}\cos{(3\theta)} } \cos{ \left[ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} \right] } + i \, e^{ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \cos{(3\theta)} } \sin{ \left[ \left( x^2 + y^2 \right) ^{\frac{3}{2}} \sin{(3\theta)} \right] }$

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