anonymous
  • anonymous
Express the complex number in trigonometric form. -3 + 3 square root of threei
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[-3+i3\sqrt3 \]
anonymous
  • anonymous
\[r = a^2 + b^2\] \[\theta =\tan^{-1}(b/a)\]
anonymous
  • anonymous
\[r^2 = 3^2 + (3\sqrt3)^2 = 36\]

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anonymous
  • anonymous
sorry, I made a typo when I was writing the formula for r, it should be: \[r =\sqrt{a^2 + b^2}\]
anonymous
  • anonymous
Anyway, we have that r = 6
anonymous
  • anonymous
We also get that theta = -60
anonymous
  • anonymous
This, however, is not correct, and highlights a flaw of the arctan function.
anonymous
  • anonymous
If we just plug in the values we got, we get this: \[6(\cos(-60) + isin(-60)) = 3 - i3\sqrt3\]
anonymous
  • anonymous
Which is actually our answer times -1
anonymous
  • anonymous
In order to rectify this, we add 180 degrees to theta.
anonymous
  • anonymous
In the end, we have: \[-3 + i3\sqrt3 = 6(\cos(120)+i \sin(120))\]
anonymous
  • anonymous
How did you get that theta=-60?

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