Question

Convert the complex number from rectangular form to polar form. z = -14 + 8i

Answer 1

`\bf \stackrel{a}{-14}+\stackrel{b}{8}i\quad \begin{cases} r[cos(\theta )+i~sin(\theta )]\\ ----------\\ r=√(a^2+b^2)\\ \theta =tan^(-1)\left( (b)/(a) \right) \end{cases} \\\\\\ r=√((-14)^2+8^2)\implies r=√(260)\implies \boxed{r=2√(65)} \\\\\\ \theta =tan^(-1)\left( (8)/(-14) \right)\implies \theta \approx -29.74^o`

now.... that angle is correct for that tangent value, however, let's take a peek at our a,b elements, "a" is negative and "b" is positive, that means the x-coordinate is negative and the y-coordinate is positive, that means, the II quadrant, NOT the IV quadrant.

so, let's get the twin of -29.74°, but adding 180° to it, and we'll land at 150.26° or thereabouts, which is in the II quadrant, where our a,b point is at.


`\bf \begin{cases} r=2√(65)\\ \theta =150.26^o \end{cases}\implies 2√(65)[cos(150.26^o)+i~sin(150.26^o)]`