Solve 5-2i/4+8i

$$\frac { 5 - 2 i } { 4 + 8 i }$$

Evaluate

$$\frac{1}{20}-\frac{3}{5}i=0.05-0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4-8i.

$$\frac{\left(5-2i\right)\left(4-8i\right)}{\left(4+8i\right)\left(4-8i\right)}$$

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

$$\frac{\left(5-2i\right)\left(4-8i\right)}{4^{2}-8^{2}i^{2}}$$

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

$$\frac{\left(5-2i\right)\left(4-8i\right)}{80}$$

Multiply complex numbers 5-2i and 4-8i like you multiply binomials.

$$\frac{5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)i^{2}}{80}$$

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

$$\frac{5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)\left(-1\right)}{80}$$

Do the multiplications in 5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)\left(-1\right).

$$\frac{20-40i-8i-16}{80}$$

Combine the real and imaginary parts in 20-40i-8i-16.

$$\frac{20-16+\left(-40-8\right)i}{80}$$

Do the additions in 20-16+\left(-40-8\right)i.

$$\frac{4-48i}{80}$$

Divide 4-48i by 80 to get \frac{1}{20}-\frac{3}{5}i.

$$\frac{1}{20}-\frac{3}{5}i$$

Real Part

$$\frac{1}{20} = 0.05$$

Solution Steps

Multiply both numerator and denominator of \frac{5-2i}{4+8i} by the complex conjugate of the denominator, 4-8i.

$$Re(\frac{\left(5-2i\right)\left(4-8i\right)}{\left(4+8i\right)\left(4-8i\right)})$$

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

$$Re(\frac{\left(5-2i\right)\left(4-8i\right)}{4^{2}-8^{2}i^{2}})$$

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

$$Re(\frac{\left(5-2i\right)\left(4-8i\right)}{80})$$

Multiply complex numbers 5-2i and 4-8i like you multiply binomials.

$$Re(\frac{5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)i^{2}}{80})$$

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

$$Re(\frac{5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)\left(-1\right)}{80})$$

Do the multiplications in 5\times 4+5\times \left(-8i\right)-2i\times 4-2\left(-8\right)\left(-1\right).

$$Re(\frac{20-40i-8i-16}{80})$$

Combine the real and imaginary parts in 20-40i-8i-16.

$$Re(\frac{20-16+\left(-40-8\right)i}{80})$$

Do the additions in 20-16+\left(-40-8\right)i.

$$Re(\frac{4-48i}{80})$$

Divide 4-48i by 80 to get \frac{1}{20}-\frac{3}{5}i.

$$Re(\frac{1}{20}-\frac{3}{5}i)$$

The real part of \frac{1}{20}-\frac{3}{5}i is \frac{1}{20}.

$$\frac{1}{20}$$