So-called simplifying, really means, "rationalizing the denominator", which is another way of saying, "getting rid of that pesky radical in the bottom"

$\bf \cfrac{4}{3-2i}\cdot \cfrac{3+2i}{3+2i}\impliedby \textit{multiplying by the conjugate of the bottom}
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\cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{4(3+2i)}{3^2-(2i)^2}\implies \cfrac{4(3+2i)}{3^2-(4i^2)}\\\\
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recall\qquad i^2=-1\\\\
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\cfrac{4(3+2i)}{3^2-(4\cdot -1)}\implies \cfrac{4(3+2i)}{9+4}\implies \cfrac{12+8i}{13}\implies \cfrac{12}{13}+\cfrac{8}{13}i$

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3 years ago

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Help please, would it be the answer choice C or D

UkoKoshka [18]

Answer:

probably C

Explanation:

5 0

3 years ago