Solve 5-i/1+i

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

Evaluate

$$2-3i$$

Solution Steps

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

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

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

$$\frac{\left(5-i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

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

Multiply complex numbers 5-i and 1-i like you multiply binomials.

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

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

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

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

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

Combine the real and imaginary parts in 5-5i-i-1.

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

Do the additions in 5-1+\left(-5-1\right)i.

$$\frac{4-6i}{2}$$

Divide 4-6i by 2 to get 2-3i.

$$2-3i$$

Real Part

$$2$$

Solution Steps

Multiply both numerator and denominator of \frac{5-i}{1+i} by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

Multiply complex numbers 5-i and 1-i like you multiply binomials.

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

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

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

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

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

Combine the real and imaginary parts in 5-5i-i-1.

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

Do the additions in 5-1+\left(-5-1\right)i.

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

Divide 4-6i by 2 to get 2-3i.

$$Re(2-3i)$$

The real part of 2-3i is 2.

$$2$$