How do you convert $\frac{1}{\sqrt{2}} - (i) (\frac{1}{\sqrt{2}})$ $1/sqrt2 - (i)(1/sqrt2)$ to polar form?

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Answer 1 · 2018-04-21T14:48:33

The polar form is $= (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $=(cos(-pi/4)+isin(-pi/4))$

The polar form of a complex number

$z = a + i b$ $z=a+ib$

is

$z = r (cos θ + i sin θ)$ $z=r(costheta+isin theta)$

Where,

$r = | z | = \sqrt{a^{2} + b^{2}}$ $r=|z|=sqrt(a^2+b^2)$

$cos θ = \frac{a}{| z |}$ $costheta=a/(|z|)$

and

$sin θ = \frac{b}{| z |}$ $sintheta=b/(|z|)$

Here, we have

$z = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$ $z=1/sqrt2-i1/sqrt2$

$| z | = \sqrt{{(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2}} = 1$ $|z|=sqrt((1/sqrt2)^2+(1/sqrt2)^2)=1$

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

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

Therefore,

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

How do you convert 1/sqrt2 - (i)(1/sqrt2)1√2−(i)(1√2)1/sqrt2 - (i)(1/sqrt2) to polar form?