Question

Express z1=4-4i in polar form

Answer 1

`\bf \stackrel{a}{4}~~\stackrel{b}{-4}~ i~~ \begin{cases} r=&√(a^2+b^2)\\ &√(4^2+(-4)^2)\\ &√(32)\\ \theta =&tan^(-1)\left( (b)/(a) \right)\\ &tan^(-1)\left( (-4)/(4) \right)\\ &tan^(-1)(-1)\\ &(7\pi )/(4) \end{cases}\implies √(32)\left[cos\left( (7\pi )/(4) \right) +i~sin\left( (7\pi )/(4) \right) \right]`

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.