How do you convert $sqrt 2 + sqrt 2i$ to polar form?

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Answer 1 · 2016-07-04T04:20:46

$(2,pi/4).$

Let $z=sqrt2+sqrt2i=x+iy$ , say.

To convert this into polar, we need these Conversion Formula $: x=rcostheta, y=rsintheta, theta in (-pi,pi], x^2+y^2=r^2$ .

We have, $x=sqrt2, y=sqrt2$ , so, $r^2=2+2=4 rArr r=2$ .

Now, $x=rcostheta rArr sqrt2=2costhetarArr costheta=1/sqrt2$ .

Similarly, $sintheta=1/sqrt2$ . Clearly, $theta=pi/4.$

Hence, the polar form of the given complex no. is $(2,pi/4).$

How do you convert sqrt 2 + sqrt 2isqrt 2 + sqrt 2i to polar form?