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))#