Solve 1|3-i/1+2i|=

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

Evaluate

$$\sqrt{2}\approx 1.414213562$$

Solution Steps

Multiply both numerator and denominator of \frac{3-i}{1+2i} by the complex conjugate of the denominator, 1-2i.

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

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

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

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

$$1|\frac{\left(3-i\right)\left(1-2i\right)}{5}|$$

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

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

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

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

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

$$1|\frac{3-6i-i-2}{5}|$$

Combine the real and imaginary parts in 3-6i-i-2.

$$1|\frac{3-2+\left(-6-1\right)i}{5}|$$

Do the additions in 3-2+\left(-6-1\right)i.

$$1|\frac{1-7i}{5}|$$

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

$$1|\frac{1}{5}-\frac{7}{5}i|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of \frac{1}{5}-\frac{7}{5}i is \sqrt{2}.

$$1\sqrt{2}$$

For any term t, t\times 1=t and 1t=t.

$$\sqrt{2}$$

Real Part

$$\sqrt{2} = 1.414213562$$