Solve 5-6i/-5+10i

$$\frac { 5 - 6 i } { - 5 + 10 i }$$

Evaluate

$$-\frac{17}{25}-\frac{4}{25}i=-0.68-0.16i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -5-10i.

$$\frac{\left(5-6i\right)\left(-5-10i\right)}{\left(-5+10i\right)\left(-5-10i\right)}$$

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

$$\frac{\left(5-6i\right)\left(-5-10i\right)}{\left(-5\right)^{2}-10^{2}i^{2}}$$

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

$$\frac{\left(5-6i\right)\left(-5-10i\right)}{125}$$

Multiply complex numbers 5-6i and -5-10i like you multiply binomials.

$$\frac{5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)i^{2}}{125}$$

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

$$\frac{5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)\left(-1\right)}{125}$$

Do the multiplications in 5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)\left(-1\right).

$$\frac{-25-50i+30i-60}{125}$$

Combine the real and imaginary parts in -25-50i+30i-60.

$$\frac{-25-60+\left(-50+30\right)i}{125}$$

Do the additions in -25-60+\left(-50+30\right)i.

$$\frac{-85-20i}{125}$$

Divide -85-20i by 125 to get -\frac{17}{25}-\frac{4}{25}i.

$$-\frac{17}{25}-\frac{4}{25}i$$

Real Part

$$-\frac{17}{25} = -0.68$$

Solution Steps

Multiply both numerator and denominator of \frac{5-6i}{-5+10i} by the complex conjugate of the denominator, -5-10i.

$$Re(\frac{\left(5-6i\right)\left(-5-10i\right)}{\left(-5+10i\right)\left(-5-10i\right)})$$

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

$$Re(\frac{\left(5-6i\right)\left(-5-10i\right)}{\left(-5\right)^{2}-10^{2}i^{2}})$$

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

$$Re(\frac{\left(5-6i\right)\left(-5-10i\right)}{125})$$

Multiply complex numbers 5-6i and -5-10i like you multiply binomials.

$$Re(\frac{5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)i^{2}}{125})$$

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

$$Re(\frac{5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)\left(-1\right)}{125})$$

Do the multiplications in 5\left(-5\right)+5\times \left(-10i\right)-6i\left(-5\right)-6\left(-10\right)\left(-1\right).

$$Re(\frac{-25-50i+30i-60}{125})$$

Combine the real and imaginary parts in -25-50i+30i-60.

$$Re(\frac{-25-60+\left(-50+30\right)i}{125})$$

Do the additions in -25-60+\left(-50+30\right)i.

$$Re(\frac{-85-20i}{125})$$

Divide -85-20i by 125 to get -\frac{17}{25}-\frac{4}{25}i.

$$Re(-\frac{17}{25}-\frac{4}{25}i)$$

The real part of -\frac{17}{25}-\frac{4}{25}i is -\frac{17}{25}.

$$-\frac{17}{25}$$