Solve 1-4i/3+i

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

Evaluate

$$-\frac{1}{10}-\frac{13}{10}i=-0.1-1.3i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-i.

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

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

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

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

$$\frac{\left(1-4i\right)\left(3-i\right)}{10}$$

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

$$\frac{1\times 3+1\left(-i\right)-4i\times 3-4\left(-1\right)i^{2}}{10}$$

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

$$\frac{1\times 3+1\left(-i\right)-4i\times 3-4\left(-1\right)\left(-1\right)}{10}$$

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

$$\frac{3-i-12i-4}{10}$$

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

$$\frac{3-4+\left(-1-12\right)i}{10}$$

Do the additions in 3-4+\left(-1-12\right)i.

$$\frac{-1-13i}{10}$$

Divide -1-13i by 10 to get -\frac{1}{10}-\frac{13}{10}i.

$$-\frac{1}{10}-\frac{13}{10}i$$

Real Part

$$-\frac{1}{10} = -0.1$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(1-4i\right)\left(3-i\right)}{10})$$

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

$$Re(\frac{1\times 3+1\left(-i\right)-4i\times 3-4\left(-1\right)i^{2}}{10})$$

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

$$Re(\frac{1\times 3+1\left(-i\right)-4i\times 3-4\left(-1\right)\left(-1\right)}{10})$$

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

$$Re(\frac{3-i-12i-4}{10})$$

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

$$Re(\frac{3-4+\left(-1-12\right)i}{10})$$

Do the additions in 3-4+\left(-1-12\right)i.

$$Re(\frac{-1-13i}{10})$$

Divide -1-13i by 10 to get -\frac{1}{10}-\frac{13}{10}i.

$$Re(-\frac{1}{10}-\frac{13}{10}i)$$

The real part of -\frac{1}{10}-\frac{13}{10}i is -\frac{1}{10}.

$$-\frac{1}{10}$$