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,

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

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