How do you multiply $e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2017-11-18T14:47:19

$e^((4pi)/3i)*e^((3pi)/2i)=cos((17pi)/2)+isin((17pi)/2)$

We can write exponential fom of complex number $e^(itheta)$ in trigonometric form as $costheta+isintheta$

Hence $e^((4pi)/3i)=cos((4pi)/3)+isin((4pi)/3)$

and $e^((3pi)/2i)=cos((3pi)/2)+isin((3pi)/2)$

Hence $e^((4pi)/3i)*e^((3pi)/2i)$

= $(cos((4pi)/3)+isin((4pi)/3))*(cos((3pi)/2)+isin((3pi)/2))$

= $cos((4pi)/3)cos((3pi)/2)+i^2sin((4pi)/3)sin((3pi)/2)+i{sin((4pi)/3)cos((3pi)/2)+cos((4pi)/3)sin((3pi)/2)}$

= ${cos((4pi)/3)cos((3pi)/2)-sin((4pi)/3)sin((3pi)/2)}+i{sin((4pi)/3)cos((3pi)/2)+cos((4pi)/3)sin((3pi)/2)}$

= $cos((4pi)/3+(3pi)/2)+isin((4pi)/3+(3pi)/2)$

= $cos((17pi)/2)+isin((17pi)/2)$

Observe that this is $e^((17pi)/2)$ which is nothing but $e^((4pi)/3+(3pi)/2)$ .

How do you multiply e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i ) e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i ) in trigonometric form?