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Convert to trigonometric form. 2 + 2i sqrt 3
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\[2 + 2i \sqrt{3}\]
http://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.577079.html hope this helps:)
indeed, it'll help him
The answer I got is (4)e^i(2/3)π
Is that correct?
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\[\Large 2+2 i \sqrt{3}=4 \left(\cos \left(\frac{\pi }{3}\right)+i \sin \left(\frac{\pi }{3}\right)\right)=4 e^{\frac{i \pi }{3}} \]
Oh man, I was way off.
\[ 2+2 i \sqrt{3}=4 \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=4 \left(\cos \left(\frac{\pi }{3}\right)+i \sin \left(\frac{\pi }{3}\right)\right)=4 e^{\frac{i \pi }{3}} \]
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