Solve 1-3i/1-i

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

Evaluate

$$2-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 4-2i by 2 to get 2-i.

$$2-i$$

Real Part

$$2$$

Solution Steps

Multiply both numerator and denominator of \frac{1-3i}{1-i} by the complex conjugate of the denominator, 1+i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 4-2i by 2 to get 2-i.

$$Re(2-i)$$

The real part of 2-i is 2.

$$2$$