Solve 2/(1+i)(3+i)

$$\frac { 2 } { ( 1 + i ) ( 3 + i ) }$$

Evaluate

$$\frac{1}{5}-\frac{2}{5}i=0.2-0.4i$$

Solution Steps

Multiply complex numbers 1+i and 3+i like you multiply binomials.

$$\frac{2}{1\times 3+i+3i+i^{2}}$$

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

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

Do the multiplications in 1\times 3+i+3i-1.

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

Combine the real and imaginary parts in 3+i+3i-1.

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

Do the additions in 3-1+\left(1+3\right)i.

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

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

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

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

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

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

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

Multiply 2 times 2-4i.

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

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

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

Divide 4-8i by 20 to get \frac{1}{5}-\frac{2}{5}i.

$$\frac{1}{5}-\frac{2}{5}i$$

Real Part

$$\frac{1}{5} = 0.2$$

Solution Steps

Multiply complex numbers 1+i and 3+i like you multiply binomials.

$$Re(\frac{2}{1\times 3+i+3i+i^{2}})$$

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

$$Re(\frac{2}{1\times 3+i+3i-1})$$

Do the multiplications in 1\times 3+i+3i-1.

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

Combine the real and imaginary parts in 3+i+3i-1.

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

Do the additions in 3-1+\left(1+3\right)i.

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

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

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

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

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

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

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

Multiply 2 times 2-4i.

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

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

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

Divide 4-8i by 20 to get \frac{1}{5}-\frac{2}{5}i.

$$Re(\frac{1}{5}-\frac{2}{5}i)$$

The real part of \frac{1}{5}-\frac{2}{5}i is \frac{1}{5}.

$$\frac{1}{5}$$