Solve 4-3i/-2+5i

$$\frac { 4 - 3 i } { - 2 + 5 i }$$

Evaluate

$$-\frac{23}{29}-\frac{14}{29}i\approx -0.793103448-0.482758621i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -2-5i.

$$\frac{\left(4-3i\right)\left(-2-5i\right)}{\left(-2+5i\right)\left(-2-5i\right)}$$

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

$$\frac{\left(4-3i\right)\left(-2-5i\right)}{\left(-2\right)^{2}-5^{2}i^{2}}$$

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

$$\frac{\left(4-3i\right)\left(-2-5i\right)}{29}$$

Multiply complex numbers 4-3i and -2-5i like you multiply binomials.

$$\frac{4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)i^{2}}{29}$$

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

$$\frac{4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)\left(-1\right)}{29}$$

Do the multiplications in 4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)\left(-1\right).

$$\frac{-8-20i+6i-15}{29}$$

Combine the real and imaginary parts in -8-20i+6i-15.

$$\frac{-8-15+\left(-20+6\right)i}{29}$$

Do the additions in -8-15+\left(-20+6\right)i.

$$\frac{-23-14i}{29}$$

Divide -23-14i by 29 to get -\frac{23}{29}-\frac{14}{29}i.

$$-\frac{23}{29}-\frac{14}{29}i$$

Real Part

$$-\frac{23}{29} = -0.7931034482758621$$

Solution Steps

Multiply both numerator and denominator of \frac{4-3i}{-2+5i} by the complex conjugate of the denominator, -2-5i.

$$Re(\frac{\left(4-3i\right)\left(-2-5i\right)}{\left(-2+5i\right)\left(-2-5i\right)})$$

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

$$Re(\frac{\left(4-3i\right)\left(-2-5i\right)}{\left(-2\right)^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{\left(4-3i\right)\left(-2-5i\right)}{29})$$

Multiply complex numbers 4-3i and -2-5i like you multiply binomials.

$$Re(\frac{4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)i^{2}}{29})$$

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

$$Re(\frac{4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)\left(-1\right)}{29})$$

Do the multiplications in 4\left(-2\right)+4\times \left(-5i\right)-3i\left(-2\right)-3\left(-5\right)\left(-1\right).

$$Re(\frac{-8-20i+6i-15}{29})$$

Combine the real and imaginary parts in -8-20i+6i-15.

$$Re(\frac{-8-15+\left(-20+6\right)i}{29})$$

Do the additions in -8-15+\left(-20+6\right)i.

$$Re(\frac{-23-14i}{29})$$

Divide -23-14i by 29 to get -\frac{23}{29}-\frac{14}{29}i.

$$Re(-\frac{23}{29}-\frac{14}{29}i)$$

The real part of -\frac{23}{29}-\frac{14}{29}i is -\frac{23}{29}.

$$-\frac{23}{29}$$