How do you convert #5 sqrt 3 - 5i# to polar form?

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Answer 1 · 2016-12-30T06:44:35

The answer is #=10(cos(-pi/6)+isin(-pi/6))#

Let #z=5sqrt3-5i#

We must transform this equation to the form

#z=r(cos theta+isintheta)#

We calculate the modulus of #z#

#∣z∣=sqrt(25*3+25)=sqrt100=10#

#z=∣z∣((5sqrt3)/(∣z∣)-i5/(∣z∣))#

#z=10(5sqrt3/10-i5/10)#

#z=10(sqrt3/2-1/2i)#

Therefore,

#costheta=sqrt3/2# and #sintheta=-1/2#

We are in the 4th quadrant

#theta=-pi/6#

Therefore,

#z=10(cos(-pi/6)+isin(-pi/6))#

#z=e^((-ipi)/6)#