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goldenfox [79]
1 year ago
15

Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Mathematics
1 answer:
Thepotemich [5.8K]1 year ago
6 0

\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

Explanation

\frac{1}{a+bi}=c+di

Step 1

multiplicate by the conjugate

\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}

notice that

\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\  \end{gathered}\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

I hope this helsp you

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