Question

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

Answer 1

`\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