1 Answer

Thaweatherman · Answer 1 · 2022-03-10T13:13:35+0000

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

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

Multiply both numerator and denominator of
$\sf (5-3i)/(-2-9i) \\$ by the complex conjugate of the denominator, -2+9i.

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

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

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

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

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

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

$\large \bf \: Re((5\left(-2\right)+5* \left(9i\right)-3i\left(-2\right)-3* 9i^(2))/(85)) \\$

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

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

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

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

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

$\large \bf Re((17+51i)/(85)) \\$

Divide 17+51i by 85 to get
$\sf(1)/(5)+(3)/(5)i \\$ .

$\large \bf \: Re((1)/(5)+(3)/(5)i) \\$

The real part of
$\sf (1)/(5)+(3)/(5)i \\$ is
$\sf (1)/(5) \\$ .

$\large \boxed{\bf(1)/(5) = 0.2} \\$

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