Solve -2-4i/-5+9i

$$\frac { - 2 - 4 i } { - 5 + 9 i }$$

Evaluate

$$-\frac{13}{53}+\frac{19}{53}i\approx -0.245283019+0.358490566i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -5-9i.

$$\frac{\left(-2-4i\right)\left(-5-9i\right)}{\left(-5+9i\right)\left(-5-9i\right)}$$

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

$$\frac{\left(-2-4i\right)\left(-5-9i\right)}{\left(-5\right)^{2}-9^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{\left(-2-4i\right)\left(-5-9i\right)}{106}$$

Multiply complex numbers -2-4i and -5-9i like you multiply binomials.

$$\frac{-2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)i^{2}}{106}$$

By definition, i^{2} is -1.

$$\frac{-2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)\left(-1\right)}{106}$$

Do the multiplications in -2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)\left(-1\right).

$$\frac{10+18i+20i-36}{106}$$

Combine the real and imaginary parts in 10+18i+20i-36.

$$\frac{10-36+\left(18+20\right)i}{106}$$

Do the additions in 10-36+\left(18+20\right)i.

$$\frac{-26+38i}{106}$$

Divide -26+38i by 106 to get -\frac{13}{53}+\frac{19}{53}i.

$$-\frac{13}{53}+\frac{19}{53}i$$

Real Part

$$-\frac{13}{53} = -0.24528301886792453$$

Solution Steps

Multiply both numerator and denominator of \frac{-2-4i}{-5+9i} by the complex conjugate of the denominator, -5-9i.

$$Re(\frac{\left(-2-4i\right)\left(-5-9i\right)}{\left(-5+9i\right)\left(-5-9i\right)})$$

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

$$Re(\frac{\left(-2-4i\right)\left(-5-9i\right)}{\left(-5\right)^{2}-9^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{\left(-2-4i\right)\left(-5-9i\right)}{106})$$

Multiply complex numbers -2-4i and -5-9i like you multiply binomials.

$$Re(\frac{-2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)i^{2}}{106})$$

By definition, i^{2} is -1.

$$Re(\frac{-2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)\left(-1\right)}{106})$$

Do the multiplications in -2\left(-5\right)-2\times \left(-9i\right)-4i\left(-5\right)-4\left(-9\right)\left(-1\right).

$$Re(\frac{10+18i+20i-36}{106})$$

Combine the real and imaginary parts in 10+18i+20i-36.

$$Re(\frac{10-36+\left(18+20\right)i}{106})$$

Do the additions in 10-36+\left(18+20\right)i.

$$Re(\frac{-26+38i}{106})$$

Divide -26+38i by 106 to get -\frac{13}{53}+\frac{19}{53}i.

$$Re(-\frac{13}{53}+\frac{19}{53}i)$$

The real part of -\frac{13}{53}+\frac{19}{53}i is -\frac{13}{53}.

$$-\frac{13}{53}$$