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anonymous

  • one year ago

Express the complex number in trigonometric form. -2 Express the complex number in trigonometric form. -2i

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  1. anonymous
    • one year ago
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    |dw:1436220030515:dw| I know you use this but how how do I find the cos. I'm a bit confused

  2. jdoe0001
    • one year ago
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    \(\large { -2\implies \begin{array}{cccllll} -2&+&0i\\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta=tan^{-1}\left( \frac{b}{a} \right)\qquad r[cos(\theta)+i\ sin(\theta)] \end{cases} }\)

  3. anonymous
    • one year ago
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    I know r is also 2 in both questions but than I'm lost from there

  4. jdoe0001
    • one year ago
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    notice who "a" and "b" is use that to get "r" or the "modulus"

  5. anonymous
    • one year ago
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    |dw:1436220473658:dw| Feel like I did this completely wrong

  6. anonymous
    • one year ago
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    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
    • one year ago
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    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. jdoe0001
    • one year ago
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    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. jdoe0001
    • one year ago
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    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
    • one year ago
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    Ahhh okay thanks soo much!

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