How do you convert #-1+i# to polar form?

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Answer 1 · 2016-05-22T12:42:09

Polar form of #-1+i# is #(sqrt2,(3pi)/4)#

A complex number #a+ib# in polar form is written as

#rcostheta+irsintheta#, #costheta=a/r# and #sintheta=b/r#

Hence #r=sqrt(a^2+b^2)#

As in #-1+i# #a=-1# and #b=1#

#r=sqrt((-1)^2+1^2)=sqrt2# and hence

#costheta=-1/sqrt2# and #sintheta=1/sqrt2#

Hence, #theta=(3pi)/4# and

Polar form of #-1+i# is #(sqrt2,(3pi)/4)#