Solve |5-i/1+i|

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

Evaluate

$$\sqrt{13}\approx 3.605551275$$

Solution Steps

Multiply both numerator and denominator of \frac{5-i}{1+i} 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.

$$|2-3i|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of 2-3i is \sqrt{13}.

$$\sqrt{13}$$

Real Part

$$\sqrt{13} = 3.605551275$$