Solve 3+i/-1+4i

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

Evaluate

$$\frac{1}{17}-\frac{13}{17}i\approx 0.058823529-0.764705882i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -1-4i.

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

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

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

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

$$\frac{\left(3+i\right)\left(-1-4i\right)}{17}$$

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

$$\frac{3\left(-1\right)+3\times \left(-4i\right)-i-4i^{2}}{17}$$

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

$$\frac{3\left(-1\right)+3\times \left(-4i\right)-i-4\left(-1\right)}{17}$$

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

$$\frac{-3-12i-i+4}{17}$$

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

$$\frac{-3+4+\left(-12-1\right)i}{17}$$

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

$$\frac{1-13i}{17}$$

Divide 1-13i by 17 to get \frac{1}{17}-\frac{13}{17}i.

$$\frac{1}{17}-\frac{13}{17}i$$

Real Part

$$\frac{1}{17} = 0.058823529411764705$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(3+i\right)\left(-1-4i\right)}{17})$$

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

$$Re(\frac{3\left(-1\right)+3\times \left(-4i\right)-i-4i^{2}}{17})$$

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

$$Re(\frac{3\left(-1\right)+3\times \left(-4i\right)-i-4\left(-1\right)}{17})$$

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

$$Re(\frac{-3-12i-i+4}{17})$$

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

$$Re(\frac{-3+4+\left(-12-1\right)i}{17})$$

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

$$Re(\frac{1-13i}{17})$$

Divide 1-13i by 17 to get \frac{1}{17}-\frac{13}{17}i.

$$Re(\frac{1}{17}-\frac{13}{17}i)$$

The real part of \frac{1}{17}-\frac{13}{17}i is \frac{1}{17}.

$$\frac{1}{17}$$