anonymous
  • anonymous
Express the complex number in trigonometric form. -2 Express the complex number in trigonometric form. -2i
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1436220030515:dw| I know you use this but how how do I find the cos. I'm a bit confused
jdoe0001
  • jdoe0001
\(\large { -2\implies \begin{array}{cccllll} -2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta=tan^{-1}\left( \frac{b}{a} \right)\qquad r[cos(\theta)+i\ sin(\theta)] \end{cases} }\)
anonymous
  • anonymous
I know r is also 2 in both questions but than I'm lost from there

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jdoe0001
  • jdoe0001
notice who "a" and "b" is use that to get "r" or the "modulus"
anonymous
  • anonymous
|dw:1436220473658:dw| Feel like I did this completely wrong
anonymous
  • 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°)
anonymous
  • 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
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
so let us see, those coordinates -2 + 0i , or -2, 0|dw:1436221647629:dw| thus \(\large { -2\implies \begin{array}{cccllll} -2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=2\\ \theta=180^o\to \pi \end{cases} \\ \quad \\ \begin{cases} 2[cos(\pi)+isin(\pi )]\\\ 2[cos(180^o)+isin(180^o)] \end{cases}}\)
anonymous
  • anonymous
Ahhh okay thanks soo much!

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