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Convert the following complex number into its polar representation: 2+2i
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Complex number to polar form: a + bi --> \(r[\cos\theta+i\sin\theta]\)
\(r=\sqrt{a^2+b^2}\)
Wouldn't square root 2^2+2^2 = 2 square root 2
\(\theta=Arctan(\large\frac{b}{a})\) if a>0 \(\theta=Arctan(\large\frac{b}{a})+\pi\) if a<0
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Yes, it would.
which is what option a. has
Yep.
oh ok i guess that would make the answer a. then but how do you figure out the stuff in the parenthesis?
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