Solve (1+i)(3+i)/3-i-(1-i)(3-i)/3+i=

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

Evaluate

$$\frac{14}{5}i=2.8i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{6+2i+12i-4}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i}$$

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

$$\frac{6-4+\left(2+12\right)i}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i}$$

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

$$\frac{2+14i}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i}$$

Divide 2+14i by 10 to get \frac{1}{5}+\frac{7}{5}i.

$$\frac{1}{5}+\frac{7}{5}i-\frac{\left(1-i\right)\left(3-i\right)}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-i^{2}\right)}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-\left(-1\right)\right)}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{3-i-3i-1}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{3-1+\left(-1-3\right)i}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{2-4i}{3+i}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{10}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{2\times 3+2\left(-i\right)-4i\times 3-4\left(-1\right)i^{2}}{10}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{2\times 3+2\left(-i\right)-4i\times 3-4\left(-1\right)\left(-1\right)}{10}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{6-2i-12i-4}{10}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{6-4+\left(-2-12\right)i}{10}$$

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

$$\frac{1}{5}+\frac{7}{5}i-\frac{2-14i}{10}$$

Divide 2-14i by 10 to get \frac{1}{5}-\frac{7}{5}i.

$$\frac{1}{5}+\frac{7}{5}i+\left(-\frac{1}{5}+\frac{7}{5}i\right)$$

Combine the real and imaginary parts in numbers \frac{1}{5}+\frac{7}{5}i and -\frac{1}{5}+\frac{7}{5}i.

$$\frac{1}{5}-\frac{1}{5}+\left(\frac{7}{5}+\frac{7}{5}\right)i$$

Add \frac{1}{5} to -\frac{1}{5}. Add \frac{7}{5} to \frac{7}{5}.

$$\frac{14}{5}i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{6+2i+12i-4}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i})$$

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

$$Re(\frac{6-4+\left(2+12\right)i}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i})$$

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

$$Re(\frac{2+14i}{10}-\frac{\left(1-i\right)\left(3-i\right)}{3+i})$$

Divide 2+14i by 10 to get \frac{1}{5}+\frac{7}{5}i.

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{\left(1-i\right)\left(3-i\right)}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-i^{2}\right)}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-\left(-1\right)\right)}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{3-i-3i-1}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{3-1+\left(-1-3\right)i}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{2-4i}{3+i})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{\left(2-4i\right)\left(3-i\right)}{10})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{2\times 3+2\left(-i\right)-4i\times 3-4\left(-1\right)i^{2}}{10})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{2\times 3+2\left(-i\right)-4i\times 3-4\left(-1\right)\left(-1\right)}{10})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{6-2i-12i-4}{10})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{6-4+\left(-2-12\right)i}{10})$$

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

$$Re(\frac{1}{5}+\frac{7}{5}i-\frac{2-14i}{10})$$

Divide 2-14i by 10 to get \frac{1}{5}-\frac{7}{5}i.

$$Re(\frac{1}{5}+\frac{7}{5}i+\left(-\frac{1}{5}+\frac{7}{5}i\right))$$

Combine the real and imaginary parts in numbers \frac{1}{5}+\frac{7}{5}i and -\frac{1}{5}+\frac{7}{5}i.

$$Re(\frac{1}{5}-\frac{1}{5}+\left(\frac{7}{5}+\frac{7}{5}\right)i)$$

Add \frac{1}{5} to -\frac{1}{5}. Add \frac{7}{5} to \frac{7}{5}.

$$Re(\frac{14}{5}i)$$

The real part of \frac{14}{5}i is 0.

$$0$$