Solve -2+5i/-3-2i

$$\frac { - 2 + 5 i } { - 3 - 2 i }$$

Evaluate

$$-\frac{4}{13}-\frac{19}{13}i\approx -0.307692308-1.461538462i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -3+2i.

$$\frac{\left(-2+5i\right)\left(-3+2i\right)}{\left(-3-2i\right)\left(-3+2i\right)}$$

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

$$\frac{\left(-2+5i\right)\left(-3+2i\right)}{\left(-3\right)^{2}-2^{2}i^{2}}$$

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

$$\frac{\left(-2+5i\right)\left(-3+2i\right)}{13}$$

Multiply complex numbers -2+5i and -3+2i like you multiply binomials.

$$\frac{-2\left(-3\right)-2\times \left(2i\right)+5i\left(-3\right)+5\times 2i^{2}}{13}$$

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

$$\frac{-2\left(-3\right)-2\times \left(2i\right)+5i\left(-3\right)+5\times 2\left(-1\right)}{13}$$

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

$$\frac{6-4i-15i-10}{13}$$

Combine the real and imaginary parts in 6-4i-15i-10.

$$\frac{6-10+\left(-4-15\right)i}{13}$$

Do the additions in 6-10+\left(-4-15\right)i.

$$\frac{-4-19i}{13}$$

Divide -4-19i by 13 to get -\frac{4}{13}-\frac{19}{13}i.

$$-\frac{4}{13}-\frac{19}{13}i$$

Real Part

$$-\frac{4}{13} = -0.3076923076923077$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(-2+5i\right)\left(-3+2i\right)}{13})$$

Multiply complex numbers -2+5i and -3+2i like you multiply binomials.

$$Re(\frac{-2\left(-3\right)-2\times \left(2i\right)+5i\left(-3\right)+5\times 2i^{2}}{13})$$

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

$$Re(\frac{-2\left(-3\right)-2\times \left(2i\right)+5i\left(-3\right)+5\times 2\left(-1\right)}{13})$$

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

$$Re(\frac{6-4i-15i-10}{13})$$

Combine the real and imaginary parts in 6-4i-15i-10.

$$Re(\frac{6-10+\left(-4-15\right)i}{13})$$

Do the additions in 6-10+\left(-4-15\right)i.

$$Re(\frac{-4-19i}{13})$$

Divide -4-19i by 13 to get -\frac{4}{13}-\frac{19}{13}i.

$$Re(-\frac{4}{13}-\frac{19}{13}i)$$

The real part of -\frac{4}{13}-\frac{19}{13}i is -\frac{4}{13}.

$$-\frac{4}{13}$$