How do you multiply e^(( 19pi )/ 8 i) * e^( pi/2 i ) e19π8i⋅eπ2i in trigonometric form?

1 Answer
Monzur R.
May 31, 2017

e^(19/8pi i)*e^(1/2pi i)=1/2(sqrt(2-sqrt2)e198πi⋅e12πi=12(√2−√2 i-sqrt(2+sqrt2))i−√2+√2)

Explanation:

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)e198πi⋅e12πi=e(198π+12π)i=e238πi

Recall Euler's formula:

e^(theta i) -=costheta+isinthetaeθi≡cosθ+isinθ

e^(23/8 pi i)=cos(23/8 pi)+isin(23/8 pi)e238πi=cos(238π)+isin(238π)

cos(23/8pi)-sqrt(2+sqrt2)/2cos(238π)−√2+√22

isin(23/8pi)=sqrt(2-sqrt2)/2 iisin(238π)=√2−√22i

Adding the two, we get:

sqrt(2-sqrt2)/2√2−√22 i-sqrt(2+sqrt2)/2=1/2(sqrt(2-sqrt2)i−√2+√22=12(√2−√2 i-sqrt(2+sqrt2))i−√2+√2)

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

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