## anonymous one year ago Express the complex number in trigonometric form. -2 Express the complex number in trigonometric form. -2i

1. anonymous

|dw:1436220030515:dw| I know you use this but how how do I find the cos. I'm a bit confused

2. anonymous

$$\large { -2\implies \begin{array}{cccllll} -2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta=tan^{-1}\left( \frac{b}{a} \right)\qquad r[cos(\theta)+i\ sin(\theta)] \end{cases} }$$

3. anonymous

I know r is also 2 in both questions but than I'm lost from there

4. anonymous

notice who "a" and "b" is use that to get "r" or the "modulus"

5. anonymous

|dw:1436220473658:dw| Feel like I did this completely wrong

6. anonymous

I don't understand how to get the cos in degrees, such as in the answer choices. Here are the answer choices: 2(cos 90° + i sin 90°) 2(cos 0° + i sin 0°) 2(cos 180° + i sin 180°) 2(cos 270° + i sin 270°)

7. anonymous

Would it be 180 degrees? I think that would be it, but not exactly sure how I really found it, its just a wild guess

8. anonymous

well.... your modulus is 2, that's correct now, for the angle, whenever either "a" or "b" is 0, you'd want to simply graph them to get the angle because range constraints on inverse funcitons, will yield non-workable values so in this case, the arcTangent function has a range of $$\textit{Inverse Trigonometric Identities} \\ \quad \\ \begin{array}{cccl} Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\ \hline\\ {\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2} \\ \quad \\ {\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi \\ \quad \\ {\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2} \end{array}$$ so... as you can see, 0 angle won't work, even though is a valid return value from the inverse tangent function

9. anonymous

so let us see, those coordinates -2 + 0i , or -2, 0|dw:1436221647629:dw| thus $$\large { -2\implies \begin{array}{cccllll} -2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=2\\ \theta=180^o\to \pi \end{cases} \\ \quad \\ \begin{cases} 2[cos(\pi)+isin(\pi )]\\\ 2[cos(180^o)+isin(180^o)] \end{cases}}$$

10. anonymous

Ahhh okay thanks soo much!