Solve 3-2i/1+4i

$$\frac { 3 - 2 i } { 1 + 4 i }$$

Evaluate

$$-\frac{5}{17}-\frac{14}{17}i\approx -0.294117647-0.823529412i$$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers 3-2i and 1-4i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\times \left(-4i\right)-2i-2\left(-4\right)\left(-1\right).

$$\frac{3-12i-2i-8}{17}$$

Combine the real and imaginary parts in 3-12i-2i-8.

$$\frac{3-8+\left(-12-2\right)i}{17}$$

Do the additions in 3-8+\left(-12-2\right)i.

$$\frac{-5-14i}{17}$$

Divide -5-14i by 17 to get -\frac{5}{17}-\frac{14}{17}i.

$$-\frac{5}{17}-\frac{14}{17}i$$

Real Part

$$-\frac{5}{17} = -0.29411764705882354$$

Solution Steps

Multiply both numerator and denominator of \frac{3-2i}{1+4i} by the complex conjugate of the denominator, 1-4i.

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

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

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

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

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

Multiply complex numbers 3-2i and 1-4i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\times \left(-4i\right)-2i-2\left(-4\right)\left(-1\right).

$$Re(\frac{3-12i-2i-8}{17})$$

Combine the real and imaginary parts in 3-12i-2i-8.

$$Re(\frac{3-8+\left(-12-2\right)i}{17})$$

Do the additions in 3-8+\left(-12-2\right)i.

$$Re(\frac{-5-14i}{17})$$

Divide -5-14i by 17 to get -\frac{5}{17}-\frac{14}{17}i.

$$Re(-\frac{5}{17}-\frac{14}{17}i)$$

The real part of -\frac{5}{17}-\frac{14}{17}i is -\frac{5}{17}.

$$-\frac{5}{17}$$