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

1 Answer
Jul 3, 2016

#-2+2i = sqrt(8)[cos((3pi)/4) + isin((3pi)/4)]#


First we want to draw an Argand diagram.

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The modulus (or length) of the complex number is simply given by using Pythagoras' theorem to find the hypotenuse, ie

#r = sqrt(x^2+y^2) = sqrt((-2)^2+2^2) = sqrt(8)#

The argument is defined as the angle from the x-axis in the counter-clockwise direction. In this case, the angle we find is in the second quadrant, so we find the 'positive' version and then need to take it away from #pi#.

#theta = pi - arctan(y/x) = pi - arctan((2)/(2)) = pi - pi/4 = (3pi)/4#