Solve 2-4i/1+3i

$$\frac { 2 - 4 i } { 1 + 3 i }$$

Evaluate

$$-1-i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-3i.

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

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

$$\frac{\left(2-4i\right)\left(1-3i\right)}{1^{2}-3^{2}i^{2}}$$

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

$$\frac{\left(2-4i\right)\left(1-3i\right)}{10}$$

Multiply complex numbers 2-4i and 1-3i like you multiply binomials.

$$\frac{2\times 1+2\times \left(-3i\right)-4i-4\left(-3\right)i^{2}}{10}$$

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

$$\frac{2\times 1+2\times \left(-3i\right)-4i-4\left(-3\right)\left(-1\right)}{10}$$

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

$$\frac{2-6i-4i-12}{10}$$

Combine the real and imaginary parts in 2-6i-4i-12.

$$\frac{2-12+\left(-6-4\right)i}{10}$$

Do the additions in 2-12+\left(-6-4\right)i.

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

Divide -10-10i by 10 to get -1-i.

$$-1-i$$

Real Part

$$-1$$

Solution Steps

Multiply both numerator and denominator of \frac{2-4i}{1+3i} by the complex conjugate of the denominator, 1-3i.

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

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

$$Re(\frac{\left(2-4i\right)\left(1-3i\right)}{1^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{\left(2-4i\right)\left(1-3i\right)}{10})$$

Multiply complex numbers 2-4i and 1-3i like you multiply binomials.

$$Re(\frac{2\times 1+2\times \left(-3i\right)-4i-4\left(-3\right)i^{2}}{10})$$

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

$$Re(\frac{2\times 1+2\times \left(-3i\right)-4i-4\left(-3\right)\left(-1\right)}{10})$$

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

$$Re(\frac{2-6i-4i-12}{10})$$

Combine the real and imaginary parts in 2-6i-4i-12.

$$Re(\frac{2-12+\left(-6-4\right)i}{10})$$

Do the additions in 2-12+\left(-6-4\right)i.

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

Divide -10-10i by 10 to get -1-i.

$$Re(-1-i)$$

The real part of -1-i is -1.

$$-1$$