# Determine all solutions to z^5 = root(3) + iAnswer in polar form.

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### 1 Answer

Put sqrt3+i=t => z^5=t => z=(t)^(1/5)

Use polar form for t:

t = r(cos theta+i sin theta)

r=sqrt((sqrt3)^2+1^2) => r=sqrt4 => r=2

Use the formula to express theta: tan theta=coefficient of i/number alone= imaginary part of complex number/real part of complex number

tan theta=1/sqrt3 => theta=30 degrees or pi/6 radians

t=2(cos(pi/6)+isin(pi/6))

z=(2)^(1/5)*(cos(pi/6)+isin(pi/6))^(1/5)

Use De Moivre's theorem:

z=(2)^(1/5)*(cos(pi/6+2npi)/5+isin(pi/6+2npi)/5)

Answer: put n=0=>z=(2)^(1/5)*(cos(pi/30)+isin(pi/30))

put n=1=>z=(2)^(1/5)*(cos(13pi/30)+isin(13pi/30))

put n=2=>z=(2)^(1/5)*(cos(25pi/30)+isin(25pi/30))

put n=3=>z=(2)^(1/5)*(cos(37pi/30)+isin(37pi/30))

put n=4=>z=(2)^(1/5)*(cos(49pi/30)+isin(49pi/30))