How do you convert #1/sqrt2 - (i)(1/sqrt2)# to polar form?

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Answer 1 · 2018-04-21T14:48:33

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

The polar form of a complex number

#z=a+ib#

is

#z=r(costheta+isin theta)#

Where,

#r=|z|=sqrt(a^2+b^2)#

#costheta=a/(|z|)#

and

#sintheta=b/(|z|)#

Here, we have

#z=1/sqrt2-i1/sqrt2#

#|z|=sqrt((1/sqrt2)^2+(1/sqrt2)^2)=1#

#{(costheta=1/sqrt2),(sintheta=-1/2):}#

#<=>#, #theta=-pi/4#, #[mod 2pi]#

Therefore,

#z=(cos(-pi/4)+isin(-pi/4))#