Question

\\(\dfrac{-4 + 9i}{2 + 7i} = \square\\)

Explanation:

: $55/53=1.037$, no, wait:
Wait, re-do numerator:
$$(-4+9i)(2-7i) = -4*2 + (-4)*(-7i) +9i*2 +9i*(-7i)$$
$$= -8 +28i +18i -63i^2$$
$$= -8 +46i +63$$
$$= 55 +46i$$
Denominator: $4 +49=53$
Wait, no, 55 divided by 53 is not 1. Wait, I made a mistake in multiplication:
Wait, $(-4)(-7i)=28i$, $9i*2=18i$, total $46i$. $-4*2=-8$, $9i*(-7i)=-63i^2=63$. $-8+63=55$. Correct.
Wait, maybe original problem was $\frac{-4+9i}{2+7i}$:
Wait, let's use another method:
Let $\frac{-4+9i}{2+7i}=a+bi$
Then $-4+9i=(a+bi)(2+7i)=2a +7ai +2bi +7bi^2=(2a-7b)+(7a+2b)i$
Set up equations:
$2a-7b=-4$
$7a+2b=9$
Solve first equation for $2a=7b-4$ → $a=\frac{7b-4}{2}$
Substitute into second equation:
$7*(\frac{7b-4}{2}) +2b=9$
Multiply by 2: $7(7b-4)+4b=18$
$49b-28+4b=18$
$53b=46$ → $b=\frac{46}{53}$
Then $a=\frac{7*(46/53)-4}{2}=\frac{322/53 -212/53}{2}=\frac{110/53}{2}=\frac{55}{53}$
So correct answer is $\frac{55}{53}+\frac{46}{53}i$
Wait, my initial miscalculation thought 55=53+2, so $1+\frac{2}{53}+\frac{46}{53}i$, no, 55/53 is 1 + 2/53. But the correct simplified form is $\frac{55}{53} + \frac{46}{53}i$

Step1: Multiply by conjugate

Multiply numerator/denominator by $2-7i$:
$$\frac{(-4+9i)(2-7i)}{(2+7i)(2-7i)}$$

Step2: Expand numerator

FOIL and use $i^2=-1$:
$$(-4)(2)+(-4)(-7i)+9i(2)+9i(-7i)=-8+28i+18i+63=55+46i$$

Step3: Expand denominator

Difference of squares, $i^2=-1$:
$$2^2-(7i)^2=4+49=53$$

Step4: Simplify fraction

Divide terms by denominator:
$$\frac{55+46i}{53}=\frac{55}{53}+\frac{46}{53}i$$

Answer:

$\frac{55}{53} + \frac{46}{53}i$