Question

Perform the indicated operations. Write the answer in standard form, a+bi. 5-3i / -2-9i

Answer 1

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i }$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i}$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)})$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}})$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85})$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85})$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$.

$\large \bf \: Re(\frac{-10+45i+6i+27}{85})$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85})$

Do the additions in $\sf-10+27+\left(45+6\right)i$.

$\large \bf Re(\frac{17+51i}{85})$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i$.

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i)$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i$ is $\sf \frac{1}{5}$.

$\large \boxed{\bf\frac{1}{5} = 0.2}$