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

User Emiko
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1 Answer

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\begin{gathered} c=(a)/(a^2+b^2) \\ d=(-b)/(a^2+b^2) \end{gathered}

Step-by-step explanation


(1)/(a+bi)=c+di

Step 1

multiplicate by the conjugate


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

notice that


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

I hope this helsp you

User Galwegian
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