Question

How do I put 5/7-i in standard form?

Answer 1

To put

`(5)/(7-i)`

In the standard form, we must multiply the numerator and denominator by the complex conjugate of (7-i), it means

`(5)/(7-i)\cdot(7+i)/(7+i)`

And now we solve it, therefore

`(5)/(7-i)\cdot(7+i)/(7+i)=(5(7+i))/(7^2-i^2)`

Remember that

`i^2=-1`

Then

`\begin{gathered} (5(7+i))/(7^2-i^2)=(5(7+i))/(49+1) \\ \\ (5(7+i))/(49+1)=(5(7+i))/(50) \end{gathered}`

Now we can simplify it

`(5(7+i))/(50)=(7+i)/(10)`

And we have it in the standard form

`(7)/(10)+(1)/(10)i`