Question

Express the complex number in trigonometric form.

3 - 3i


radians pls

Answer 1

`\bf \begin{array}{llll} 3&-&3i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=√(a^2+b^2)\\ \theta=tan^(-1)\left( (b)/(a) \right) \end{cases}\\\\ -----------------------------\\\\ r=√(3^2+3^2)\implies r=√(18)\implies r=3√(2) \\\\\\ tan(\theta)=\cfrac{3}{3}\implies tan(\theta)=1\implies \measuredangle \theta=tan^(-1)(1) \\\\\\ \measuredangle \theta=(\pi )/(4)\impliedby \textit{reference angle}`

now, b is -3, and a is 3
or in rectangular if you wish, y = -3 and x = 3.... on what quadrant is "y" negative and "x" positive? well, the 4th quadrant
so, using our reference angle, that'd be
`\bf \cfrac{7\pi }{4}\quad or \quad -\cfrac{\pi }{4}`

so, let's use the first one


`\bf (3,3i)\implies \begin{array}{llll} r[cos(\theta)+i\ sin(\theta)]\\\\ 3√(2)\left[ cos\left( (7\pi )/(4) \right)+i\ sin\left( (7\pi )/(4) \right) \right] \end{array}`