Solve 1+8i/2+6i

$$\frac { 1 + 8 i } { 2 + 6 i }$$

Evaluate

$$\frac{5}{4}+\frac{1}{4}i=1.25+0.25i$$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers 1+8i and 2-6i like you multiply binomials.

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

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

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

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

$$\frac{2-6i+16i+48}{40}$$

Combine the real and imaginary parts in 2-6i+16i+48.

$$\frac{2+48+\left(-6+16\right)i}{40}$$

Do the additions in 2+48+\left(-6+16\right)i.

$$\frac{50+10i}{40}$$

Divide 50+10i by 40 to get \frac{5}{4}+\frac{1}{4}i.

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

Real Part

$$\frac{5}{4} = 1\frac{1}{4} = 1.25$$

Solution Steps

Multiply both numerator and denominator of \frac{1+8i}{2+6i} by the complex conjugate of the denominator, 2-6i.

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

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

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

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

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

Multiply complex numbers 1+8i and 2-6i like you multiply binomials.

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

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

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

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

$$Re(\frac{2-6i+16i+48}{40})$$

Combine the real and imaginary parts in 2-6i+16i+48.

$$Re(\frac{2+48+\left(-6+16\right)i}{40})$$

Do the additions in 2+48+\left(-6+16\right)i.

$$Re(\frac{50+10i}{40})$$

Divide 50+10i by 40 to get \frac{5}{4}+\frac{1}{4}i.

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

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

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