How do you simplify $(8+6i)-(2+3i)$ ?

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Answer 1 · 2016-10-23T08:25:23

$3*sqrt(5)*(cos(0.464)+isin(0.464))$

First simplify Cartesian form,
$(8+6i)-(2+3i)= 8-2 +6i-3i$
$=6+3i$

now to convert to trigonometric form,

$6+3i = r*cis(theta)$

where,
$r=sqrt(6^2+3^2)$
$r= sqrt(45)$ (which can be further simplified to $3*sqrt(5)$ )

and,

$theta=tan^-1(3/6)$

$theta = 0.464$ (3dp, radians)

so we can express $(8+6i)-(2+3i)$ as,

$3*sqrt(5)*cis(0.464)$

or in expanded version,
$3*sqrt(5)*(cos(0.464)+isin(0.464))$

How do you simplify (8+6i)-(2+3i)(8+6i)-(2+3i)?