How do you convert #sqrt3 - sqrt3i# to polar form?

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Answer 1 · 2016-10-22T11:34:16

The polar form is #sqrt3(cos(-pi/4)+isin(-pi/4))#

Let #z=sqrt3-isqrt3#

Then the modulus of z is #∣z∣=sqrt((sqrt3)^2+ (sqrt3)^2) =sqrt6#
Rewriting z

#z=∣z∣(sqrt3-isqrt3)=sqrt6((sqrt3/sqrt6)-i(sqrt3/sqrt6))#

#=sqrt6(1/sqrt2-i/sqrt2)#

Comparing this to the standard form #z=r(costheta+isintheta)#

We see that #costheta=1/sqrt2#

and #sintheta=-1/sqrt2#
This can occur in the 4th quadrant and #theta =-pi/4#

And finally we have #z=sqrt6(cos(-pi/4)+sin(-pi/4))#

You can also write the resultas #z=sqrt6e^(-ipi/4)#