How do you multiply $e^(( 19pi )/ 8 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2017-05-31T14:02:02

$e^(19/8pi i)*e^(1/2pi i)=1/2(sqrt(2-sqrt2)$ $i-sqrt(2+sqrt2))$

By the multiplication rule of exponents, we can rewrite the equation as one exponent:

$e^(19/8pi i)*e^(1/2pi i)=e^((19/8pi+1/2pi)i)=e^(23/8pi i)$

Recall Euler's formula:

$e^(theta i) -=costheta+isintheta$

$e^(23/8 pi i)=cos(23/8 pi)+isin(23/8 pi)$

$cos(23/8pi)-sqrt(2+sqrt2)/2$

$isin(23/8pi)=sqrt(2-sqrt2)/2 i$

Adding the two, we get:

$sqrt(2-sqrt2)/2$ $i-sqrt(2+sqrt2)/2=1/2(sqrt(2-sqrt2)$ $i-sqrt(2+sqrt2))$

$therefore e^(19/8pi i)*e^(1/2pii)=1/2(sqrt(2-sqrt2)$ $i-sqrt(2+sqrt2))$

How do you multiply e^(( 19pi )/ 8 i) * e^( pi/2 i ) e^(( 19pi )/ 8 i) * e^( pi/2 i ) in trigonometric form?