Question

Simplify the following quotient of complex numbers into the form a + bi. (-4-3i)/(2-i)

Answer 1

multiply top and bottom by conjugate of the denominator
conjugate of \(a+bi\) is \(a-bi\) and this works because
\((a+bi)(a-bi)=a^2+b^2\) a real number

Answer 2

in your case you will get
\[\frac{-4-3i}{2-i}\times \frac{2+i}{2+i}=\frac{-(4-3i)(2+i)}{2^2+1^2}\]

Answer 3

typo there
\[\frac{-4-3i}{2-i}\times \frac{2+i}{2+i}=\frac{(-4-3i)(2+i)}{2^2+1^2}\]

Answer 4

denominator is 5, numerator is what you get when you multiply
\((-4-3i)(2+i)\) and btw this should be called "divide" not "simplify" just thought i would mention it

Answer 5

so the answer would be -11-10x/5 right

Answer 6

the other way is to convert to the numerator and denominator \(re^{i\theta}\)
then division is simple