How do you convert #sqrt3 - 3i# to polar form?

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Answer 1 · 2018-03-21T15:45:12

The answer is #=sqrt12(cos(-1/3pi)+isin(-1/3pi))#

Any complex number

#z=a+ib#

can be converted to the polar form

#z=|z|(costheta+i sintheta)#

Where,

#costheta=a/(|z|)#

and

#sintheta=b/(|z|)#

Here,

#z=sqrt3-3i#

#|z|=sqrt((sqrt3)^2+(-3)^2)=sqrt12#

#costheta=sqrt3/sqrt12=1/2#

#sintheta=-3/sqrt12=-sqrt3/(2)#

Therefore,

#theta=-pi/3#, #[mod 2pi]#

#z=sqrt12(cos(-1/3pi)+isin(-1/3pi))#