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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436220030515:dw I know you use this but how how do I find the cos. I'm a bit confused

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\large { 2\implies \begin{array}{cccllll} 2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta=tan^{1}\left( \frac{b}{a} \right)\qquad r[cos(\theta)+i\ sin(\theta)] \end{cases} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know r is also 2 in both questions but than I'm lost from there

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1notice who "a" and "b" is use that to get "r" or the "modulus"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436220473658:dw Feel like I did this completely wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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°)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be 180 degrees? I think that would be it, but not exactly sure how I really found it, its just a wild guess

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well.... 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 nonworkable 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

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1so let us see, those coordinates 2 + 0i , or 2, 0dw:1436221647629:dw thus \(\large { 2\implies \begin{array}{cccllll} 2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=2\\ \theta=180^o\to \pi \end{cases} \\ \quad \\ \begin{cases} 2[cos(\pi)+isin(\pi )]\\\ 2[cos(180^o)+isin(180^o)] \end{cases}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh okay thanks soo much!
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