Solve 5+i/-6-2i

$$\frac { 5 + i } { - 6 - 2 i }$$

Evaluate

$$-\frac{4}{5}+\frac{1}{10}i=-0.8+0.1i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -6+2i.

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

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

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

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

$$\frac{\left(5+i\right)\left(-6+2i\right)}{40}$$

Multiply complex numbers 5+i and -6+2i like you multiply binomials.

$$\frac{5\left(-6\right)+5\times \left(2i\right)-6i+2i^{2}}{40}$$

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

$$\frac{5\left(-6\right)+5\times \left(2i\right)-6i+2\left(-1\right)}{40}$$

Do the multiplications in 5\left(-6\right)+5\times \left(2i\right)-6i+2\left(-1\right).

$$\frac{-30+10i-6i-2}{40}$$

Combine the real and imaginary parts in -30+10i-6i-2.

$$\frac{-30-2+\left(10-6\right)i}{40}$$

Do the additions in -30-2+\left(10-6\right)i.

$$\frac{-32+4i}{40}$$

Divide -32+4i by 40 to get -\frac{4}{5}+\frac{1}{10}i.

$$-\frac{4}{5}+\frac{1}{10}i$$

Real Part

$$-\frac{4}{5} = -0.8$$

Solution Steps

Multiply both numerator and denominator of \frac{5+i}{-6-2i} by the complex conjugate of the denominator, -6+2i.

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

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

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

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

$$Re(\frac{\left(5+i\right)\left(-6+2i\right)}{40})$$

Multiply complex numbers 5+i and -6+2i like you multiply binomials.

$$Re(\frac{5\left(-6\right)+5\times \left(2i\right)-6i+2i^{2}}{40})$$

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

$$Re(\frac{5\left(-6\right)+5\times \left(2i\right)-6i+2\left(-1\right)}{40})$$

Do the multiplications in 5\left(-6\right)+5\times \left(2i\right)-6i+2\left(-1\right).

$$Re(\frac{-30+10i-6i-2}{40})$$

Combine the real and imaginary parts in -30+10i-6i-2.

$$Re(\frac{-30-2+\left(10-6\right)i}{40})$$

Do the additions in -30-2+\left(10-6\right)i.

$$Re(\frac{-32+4i}{40})$$

Divide -32+4i by 40 to get -\frac{4}{5}+\frac{1}{10}i.

$$Re(-\frac{4}{5}+\frac{1}{10}i)$$

The real part of -\frac{4}{5}+\frac{1}{10}i is -\frac{4}{5}.

$$-\frac{4}{5}$$