Solve 4+2i/1+3i

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

Evaluate

$$1-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Do the additions in 4+6+\left(-12+2\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{4+2i}{1+3i} by the complex conjugate of the denominator, 1-3i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$1$$