[tex]\bf \stackrel{a}{4}~~\stackrel{b}{-4}~ i~~ \begin{cases} r=&\sqrt{a^2+b^2}\\ &\sqrt{4^2+(-4)^2}\\ &\sqrt{32}\\ \theta =&tan^{-1}\left( \frac{b}{a} \right)\\ &tan^{-1}\left( \frac{-4}{4} \right)\\ &tan^{-1}(-1)\\ &\frac{7\pi }{4} \end{cases}\implies \sqrt{32}\left[cos\left( \frac{7\pi }{4} \right) +i~sin\left( \frac{7\pi }{4} \right) \right][/tex]
now, le't bear in mind that tan⁻¹(-1) also yields 3π/4, however, let's notice, a = 4, b = -4, meaning the x = 4, y = -4, and that only happens on the IV Quadrant, thus the angle for that tan⁻¹(-1) must be in the IV Quadrant, thus 7π/4.
