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-2+2i / 5+3i please help me simplify
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do you know what a conjugate is?
yes
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 \(\bf \cfrac{-2+2i }{5+3i }\cdot \cfrac{5-3i}{5-3i}\qquad recall\implies {\color{blue}{ a^2-b^2 = (a-b)(a+b)}}\qquad thus\\ \quad \\ \cfrac{-2+2i }{5+3i }\cdot \cfrac{5-3i}{5-3i}\implies \cfrac{(-2+2i)(5-3i) }{5^2-(3i)^2 }\implies \cfrac{(-2+2i)(5-3i) }{25-3^2{\color{blue}{ i^2}} }\\ \quad \\ \cfrac{(-2+2i)(5-3i) }{25-(9\cdot {\color{blue}{ -1}}) }\)
okay , thanks. I understand now.
yw
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