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Mathematics 16 Online
OpenStudy (anonymous):

-2+2i / 5+3i please help me simplify

OpenStudy (jdoe0001):

do you know what a conjugate is?

OpenStudy (anonymous):

yes

OpenStudy (jdoe0001):

ok... for a complex rational, "simplify" just means, get rid of the pesky "i" in the bottom so the way you'd do it is by multiplying both, top and bottom, by the denominator's conjugate, so \(\bf \cfrac{-2+2i }{5+3i }\cdot \cfrac{5-3i}{5-3i}\qquad recall\implies {\color{blue}{ a^2-b^2 = (a-b)(a+b)}}\qquad thus\\ \quad \\ \cfrac{-2+2i }{5+3i }\cdot \cfrac{5-3i}{5-3i}\implies \cfrac{(-2+2i)(5-3i) }{5^2-(3i)^2 }\implies \cfrac{(-2+2i)(5-3i) }{25-3^2{\color{blue}{ i^2}} }\\ \quad \\ \cfrac{(-2+2i)(5-3i) }{25-(9\cdot {\color{blue}{ -1}}) }\)

OpenStudy (anonymous):

okay , thanks. I understand now.

OpenStudy (jdoe0001):

yw

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