[tex]\bf \stackrel{a}{2}\qquad \stackrel{b}{-2}i~~ \begin{cases} r=&\sqrt{a^2+b^2}\\ &\sqrt{2^2+(-2)^2}\\ &\sqrt{2^2(1+1)}\\ &2\sqrt{2}\\ \theta =&tan^{-1}\left( \frac{b}{a} \right)\\ &tan^{-1}\left( \frac{-2}{2} \right)\\ &tan^{-1}(-1)\\ &\frac{3\pi }{4}~~,~~\frac{7\pi }{4}\\ &\frac{7\pi }{4} \end{cases}~\hfill \implies ~\hfill \begin{array}{llll} r[\cos(\theta )+i\sin(\theta )]\\\\ 2\sqrt{2}\left[ \cos\left( \frac{7\pi }{4} \right)+i\sin\left( \frac{7\pi }{4} \right) \right] \end{array}[/tex]
tis noteworthy that, on the range of [0 , 2π] we get two angles whose tangent is -1, one in the II Quadrant and another on the IV Quadrant, however, let's notice our a,b coordinates, "a" is positive and "b" is negative, that means the angle in the IV Quadrant, so is not 3π/4, is 7π/4 because that is the angle in the IV Quadrant.
