How do you multiply $e^(( 9 pi )/ 8 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-12-30T11:50:20

$e^((9pi)/8i) * e^(pi/2i)=color(green)(sin(pi/8)-icos(pi/8))$

Using Euler's formula
$color(white)("XXX")e^((9pi)/8i) = cos((9pi)/8)+i * sin((9pi)/8)$

$color(white)("XXXXX")=cos(pi+pi/8)+i * sin(pi+pi/8)$

$color(white)("XXXXX")=-cos(pi/8)+i * (-sin(pi/8))$

$color(white)("XXXXX")=-cos(pi/8)- i * sin(pi/8)$

and
$color(white)("XXX")e^(pi/2i)=cos(pi/2)+ i * sin(pi/2)$

$color(white)("XXXXX")=0 + i * 1$

$color(white)("XXXXX")=i$

Therefore
$color(white)("XXX")e^((9pi)/8i) * e^(pi/2i) = [-cos(pi/8)- i * sin(pi/8)] * i$

$color(white)("XXXXXXX")=- i * cos(pi/8) - (-1) * sin(pi/8)$

$color(white)("XXXXXXX")=sin(pi/8)-icos(pi/8)$

How do you multiply e^(( 9 pi )/ 8 i) * e^( pi/2 i ) e^(( 9 pi )/ 8 i) * e^( pi/2 i ) in trigonometric form?