Solve 2-4i/2+4i

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

Evaluate

$$-\frac{3}{5}-\frac{4}{5}i=-0.6-0.8i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-4i.

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

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

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

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

$$\frac{\left(2-4i\right)\left(2-4i\right)}{20}$$

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

$$\frac{2\times 2+2\times \left(-4i\right)-4i\times 2-4\left(-4\right)i^{2}}{20}$$

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

$$\frac{2\times 2+2\times \left(-4i\right)-4i\times 2-4\left(-4\right)\left(-1\right)}{20}$$

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

$$\frac{4-8i-8i-16}{20}$$

Combine the real and imaginary parts in 4-8i-8i-16.

$$\frac{4-16+\left(-8-8\right)i}{20}$$

Do the additions in 4-16+\left(-8-8\right)i.

$$\frac{-12-16i}{20}$$

Divide -12-16i by 20 to get -\frac{3}{5}-\frac{4}{5}i.

$$-\frac{3}{5}-\frac{4}{5}i$$

Real Part

$$-\frac{3}{5} = -0.6$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(2-4i\right)\left(2-4i\right)}{20})$$

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

$$Re(\frac{2\times 2+2\times \left(-4i\right)-4i\times 2-4\left(-4\right)i^{2}}{20})$$

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

$$Re(\frac{2\times 2+2\times \left(-4i\right)-4i\times 2-4\left(-4\right)\left(-1\right)}{20})$$

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

$$Re(\frac{4-8i-8i-16}{20})$$

Combine the real and imaginary parts in 4-8i-8i-16.

$$Re(\frac{4-16+\left(-8-8\right)i}{20})$$

Do the additions in 4-16+\left(-8-8\right)i.

$$Re(\frac{-12-16i}{20})$$

Divide -12-16i by 20 to get -\frac{3}{5}-\frac{4}{5}i.

$$Re(-\frac{3}{5}-\frac{4}{5}i)$$

The real part of -\frac{3}{5}-\frac{4}{5}i is -\frac{3}{5}.

$$-\frac{3}{5}$$