Question

Find the quotient of these complex numbers.(6 - i) ÷ (4 + 3i) =

Answer 1

To solve the exercise you can apply the complex arithmetic rule, that is

`(a+bi)/(c+di)=((c-di)(a+bi))/((c-di)(c+di))=((ac+bd)+(bc-ad)i)/(c^2+d^2)`

So, in this case, you have

`\begin{gathered} (6-i)/(4+3i) \\ a=6 \\ b=-1 \\ c=4 \\ d=3 \end{gathered}`
`(6-i)/(4+3i)=((6\cdot4+(-1)\cdot3)+(-1\cdot4-6\cdot3)i)/(4^2+3^2)`

Simplifying

`\begin{gathered} (6-i)/(4+3i)=((24-3)+(-4-18)i)/(16+9) \\ (6-i)/(4+3i)=(21-22i)/(16+9) \\ (6-i)/(4+3i)=(21-22i)/(25) \end{gathered}`

Finally, rewrite in binomic form, that is, z = a+bi

`(21-22i)/(25)=(21)/(25)-(22i)/(25)`

Therefore, the correct answer is B.

`(21)/(25)-(22i)/(25)`