How do you multiply $e^{\frac{π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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How do you multiply $e^{\frac{π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-03-02T07:04:09

The answer is $= \frac{\sqrt{2}}{2} (1 - i)$ $=sqrt2/2(1-i)$

Explanation:

Apply Euler's Identity

$e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$

$i^{2} = - 1$ $i^2=-1$

Therefore,

$e^{\frac{π}{4} i} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} (1 + i)$ $e^(pi/4i)=cos(pi/4)+isin(pi/4)=sqrt2/2+isqrt2/2=sqrt2/2(1+i)$

$e^{\frac{3}{2} π i} = cos (\frac{3}{2} π) + i sin (\frac{3}{2} π) = 0 - i$ $e^(3/2pii)=cos(3/2pi)+isin(3/2pi)=0-i$

So,

$z = e^{\frac{π}{4} i} \cdot e^{\frac{3}{2} π i} = \frac{\sqrt{2}}{2} (1 + i) \cdot (- i) = \frac{\sqrt{2}}{2} (- i - i^{2})$ $z=e^(pi/4i)*e^(3/2pii)=sqrt2/2(1+i)*(-i)=sqrt2/2(-i-i^2)$

$= \frac{\sqrt{2}}{2} (1 - i)$ $=sqrt2/2(1-i)$

$Verification$ $"Verification"$

$z = (cos ϕ + i sin ϕ) = \frac{\sqrt{2}}{2} (1 - i)$ $z=(cosphi+isinphi)=sqrt2/2(1-i)$

$cos ϕ = \frac{\sqrt{2}}{2}$ $cosphi=sqrt2/2$

$sin ϕ = - \frac{\sqrt{2}}{2}$ $sinphi=-sqrt2/2$

$ϕ = - \frac{π}{4}$ $phi=-pi/4$ , $[2 π]$ $[2pi]$

$z = e^{\frac{π}{4} i} \cdot e^{\frac{3}{2} π i} = e^{(\frac{π}{4} + \frac{3}{2} π) i} = e^{\frac{7}{4} π i} = e^{- \frac{1}{4} π i}$ $z=e^(pi/4i)*e^(3/2pii)=e^((pi/4+3/2pi)i)=e^(7/4pii)=e^(-1/4pii)$

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How do you multiply $e^{\frac{π}{4} i} \cdot e^{3 \frac{π}{2} i}$ $e^(( pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Explanation:

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