How do you write $6 + 0 i$ $6+0i$ in trigonometric form?

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Answer 1 · 2016-02-11T08:32:58

$6 (cos 0 + i \cdot sin 0)$ $6(cos 0+i*sin 0)$

Trigonometric form of $a + b i$ $a+bi$ is $r (cos θ + sin θ)$ $r(cos theta+sin theta)$

As such $r$ $r$ is $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ and $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^(-1)(b/a)$

Here $a = 6$ $a=6$ and $b = 0$ $b=0$ , hence

$r = \sqrt{6^{2} + 0^{2}} = 6$ $r=sqrt(6^2+0^2) = 6$ and $θ = {tan}^{- 1} 0 = 0$ $theta=tan^(-1)0=0$

Hence $6 + 0 i$ $6+0i$ in trigonometric form is $6 (cos 0 + i \cdot sin 0)$ $6(cos 0+i*sin 0)$ .

How do you write 6+0i6+0i in trigonometric form?