Solve |1+3i/-4-3i|

$$| \frac { 1 + 3 i } { - 4 - 3 i } |$$

Evaluate

$$\frac{\sqrt{10}}{5}\approx 0.632455532$$

Solution Steps

Multiply both numerator and denominator of \frac{1+3i}{-4-3i} by the complex conjugate of the denominator, -4+3i.

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

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

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

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

$$|\frac{\left(1+3i\right)\left(-4+3i\right)}{25}|$$

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

$$|\frac{1\left(-4\right)+1\times \left(3i\right)+3i\left(-4\right)+3\times 3i^{2}}{25}|$$

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

$$|\frac{1\left(-4\right)+1\times \left(3i\right)+3i\left(-4\right)+3\times 3\left(-1\right)}{25}|$$

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

$$|\frac{-4+3i-12i-9}{25}|$$

Combine the real and imaginary parts in -4+3i-12i-9.

$$|\frac{-4-9+\left(3-12\right)i}{25}|$$

Do the additions in -4-9+\left(3-12\right)i.

$$|\frac{-13-9i}{25}|$$

Divide -13-9i by 25 to get -\frac{13}{25}-\frac{9}{25}i.

$$|-\frac{13}{25}-\frac{9}{25}i|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of -\frac{13}{25}-\frac{9}{25}i is \sqrt{\frac{2}{5}}.

$$\sqrt{\frac{2}{5}}$$

Rewrite the square root of the division \sqrt{\frac{2}{5}} as the division of square roots \frac{\sqrt{2}}{\sqrt{5}}.

$$\frac{\sqrt{2}}{\sqrt{5}}$$

Rationalize the denominator of \frac{\sqrt{2}}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$\frac{\sqrt{2}\sqrt{5}}{\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{2}\sqrt{5}}{5}$$

To multiply \sqrt{2} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{\sqrt{10}}{5}$$

Real Part

$$\frac{\sqrt{10}}{5} = 0.6324555320336759$$