Question

The complex number z = –1 + 3i can be expressed in trigonometric form as:

Answer 1

`\bf z=\stackrel{a}{-1}\stackrel{b}{+3}i\qquad \begin{cases} r=√(a^2+b^2)\\ \theta =tan^(-1)\left( (b)/(a) \right) \end{cases}\qquad \qquad z=r[cos(\theta )i~sin(\theta )]\\\\ -------------------------------\\\\ r=√((-1)^2+3^2)\implies r=√(10) \\\\\\ \theta =tan^(-1)\left( (3)/(-1) \right)\implies \theta \approx -71.57\implies \theta \approx\stackrel{360-71.57}{288.43^o}\\\\ -------------------------------\\\\ z=√(10)\left[ cos(288.43^o)+i~sin(288.43^o) \right]`

now, the calculation in the choices, show 289.5°, which I take it is due some early rounding up.