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

1 Answer
Ratnaker Mehta
Jul 4, 2016

#(2,pi/4).#

Explanation:

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).#

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