Solve left(-2+8iright)divleft(-2+6iright)

$$\left( -2+8i \right) \div \left( -2+6i \right)$$

Evaluate

$$\frac{13}{10}-\frac{1}{10}i=1.3-0.1i$$

Solution Steps

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

$$\frac{\left(-2+8i\right)\left(-2-6i\right)}{\left(-2+6i\right)\left(-2-6i\right)}$$

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

$$\frac{\left(-2+8i\right)\left(-2-6i\right)}{\left(-2\right)^{2}-6^{2}i^{2}}$$

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

$$\frac{\left(-2+8i\right)\left(-2-6i\right)}{40}$$

Multiply complex numbers -2+8i and -2-6i like you multiply binomials.

$$\frac{-2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)i^{2}}{40}$$

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

$$\frac{-2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)\left(-1\right)}{40}$$

Do the multiplications in -2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)\left(-1\right).

$$\frac{4+12i-16i+48}{40}$$

Combine the real and imaginary parts in 4+12i-16i+48.

$$\frac{4+48+\left(12-16\right)i}{40}$$

Do the additions in 4+48+\left(12-16\right)i.

$$\frac{52-4i}{40}$$

Divide 52-4i by 40 to get \frac{13}{10}-\frac{1}{10}i.

$$\frac{13}{10}-\frac{1}{10}i$$

Real Part

$$\frac{13}{10} = 1\frac{3}{10} = 1.3$$

Solution Steps

Multiply both numerator and denominator of \frac{-2+8i}{-2+6i} by the complex conjugate of the denominator, -2-6i.

$$Re(\frac{\left(-2+8i\right)\left(-2-6i\right)}{\left(-2+6i\right)\left(-2-6i\right)})$$

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

$$Re(\frac{\left(-2+8i\right)\left(-2-6i\right)}{\left(-2\right)^{2}-6^{2}i^{2}})$$

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

$$Re(\frac{\left(-2+8i\right)\left(-2-6i\right)}{40})$$

Multiply complex numbers -2+8i and -2-6i like you multiply binomials.

$$Re(\frac{-2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)i^{2}}{40})$$

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

$$Re(\frac{-2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)\left(-1\right)}{40})$$

Do the multiplications in -2\left(-2\right)-2\times \left(-6i\right)+8i\left(-2\right)+8\left(-6\right)\left(-1\right).

$$Re(\frac{4+12i-16i+48}{40})$$

Combine the real and imaginary parts in 4+12i-16i+48.

$$Re(\frac{4+48+\left(12-16\right)i}{40})$$

Do the additions in 4+48+\left(12-16\right)i.

$$Re(\frac{52-4i}{40})$$

Divide 52-4i by 40 to get \frac{13}{10}-\frac{1}{10}i.

$$Re(\frac{13}{10}-\frac{1}{10}i)$$

The real part of \frac{13}{10}-\frac{1}{10}i is \frac{13}{10}.

$$\frac{13}{10}$$