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

1 Answer
Apr 21, 2018

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

Explanation:

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))