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I am unsure if I got the correct real and imaginary parts.

$e^{3} * e^{-j\frac{\pi}{4}} = e^{3} (\cos(\frac{\pi}{4}) -j\sin(\frac{\pi}{4}))$.

Is the real part then: $e^{3}\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}e^{3} $

Then the imaginary part is: $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

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  • $\begingroup$ The imaginary part is $-e^3\frac{\sqrt 2}{2}$. Do you see why? $\endgroup$ – cip999 Sep 24 '17 at 19:36
  • $\begingroup$ what is $j$ ? in math context should be primitive root of $j^3=1$, but you seem to use electricity notation for $i$. $\endgroup$ – zwim Sep 24 '17 at 19:41
  • $\begingroup$ $j$ could also be one of the quaternionic imaginaries. They are usually called $i,j,k$ $\endgroup$ – mathreadler Sep 24 '17 at 19:46
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$$ e^3\left(\frac{\sqrt{2}}{2}-j\frac{\sqrt{2}}{2} \right)=\frac{\sqrt{2}}{2} e^3 -j \frac{\sqrt{2}}{2}e^3 $$

so the real part is $\frac{\sqrt{2}}{2} e^3 $ as you found , but the imaginary part is $-\frac{\sqrt{2}}{2} e^3 $

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$${ e }^{ 3 }\cdot { e }^{ -i\frac { \pi }{ 4 } }={ e }^{ 3 }\left( \cos { \left( -\frac { \pi }{ 4 } \right) +i\sin { \left( -\frac { \pi }{ 4 } \right) } } \right) ={ e }^{ 3 }\cos { \left( \frac { \pi }{ 4 } \right) } +i{ e }^{ 3 }\sin { \left( -\frac { \pi }{ 4 } \right) } \\ Re\left( { e }^{ 3 }\cdot { e }^{ -i\frac { \pi }{ 4 } } \right) ={ e }^{ 3 }\cos { \left( \frac { \pi }{ 4 } \right) } =\frac { { e }^{ 3 } }{ \sqrt { 2 } } \\ Im\left( { e }^{ 3 }\cdot { e }^{ -i\frac { \pi }{ 4 } } \right) ={ e }^{ 3 }\sin { \left( -\frac { \pi }{ 4 } \right) } =-\frac { { e }^{ 3 } }{ \sqrt { 2 } } $$

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$$ e^{3}e^{-j\frac{\pi}{4}}=e^{3}(\cos(-\pi/4)+j\sin(-\pi/4))=e^{3}\frac{\sqrt{2}}{2}-je^{3}\frac{\sqrt{2}}{2}. $$ Real part: $e^{3}\frac{\sqrt{2}}{2}$

Imaginary part: $-e^{3}\frac{\sqrt{2}}{2}$ (you forgot the minus sign and the $e^{3}$ factor).

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