Solve 4-18i/1+4i

$$\frac { 4 - 18 i } { 1 + 4 i }$$

Evaluate

$$-4-2i$$

Solution Steps

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

$$\frac{\left(4-18i\right)\left(1-4i\right)}{\left(1+4i\right)\left(1-4i\right)}$$

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

$$\frac{\left(4-18i\right)\left(1-4i\right)}{1^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(4-18i\right)\left(1-4i\right)}{17}$$

Multiply complex numbers 4-18i and 1-4i like you multiply binomials.

$$\frac{4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)i^{2}}{17}$$

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

$$\frac{4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)\left(-1\right)}{17}$$

Do the multiplications in 4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)\left(-1\right).

$$\frac{4-16i-18i-72}{17}$$

Combine the real and imaginary parts in 4-16i-18i-72.

$$\frac{4-72+\left(-16-18\right)i}{17}$$

Do the additions in 4-72+\left(-16-18\right)i.

$$\frac{-68-34i}{17}$$

Divide -68-34i by 17 to get -4-2i.

$$-4-2i$$

Real Part

$$-4$$

Solution Steps

Multiply both numerator and denominator of \frac{4-18i}{1+4i} by the complex conjugate of the denominator, 1-4i.

$$Re(\frac{\left(4-18i\right)\left(1-4i\right)}{\left(1+4i\right)\left(1-4i\right)})$$

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

$$Re(\frac{\left(4-18i\right)\left(1-4i\right)}{1^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(4-18i\right)\left(1-4i\right)}{17})$$

Multiply complex numbers 4-18i and 1-4i like you multiply binomials.

$$Re(\frac{4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)i^{2}}{17})$$

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

$$Re(\frac{4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)\left(-1\right)}{17})$$

Do the multiplications in 4\times 1+4\times \left(-4i\right)-18i-18\left(-4\right)\left(-1\right).

$$Re(\frac{4-16i-18i-72}{17})$$

Combine the real and imaginary parts in 4-16i-18i-72.

$$Re(\frac{4-72+\left(-16-18\right)i}{17})$$

Do the additions in 4-72+\left(-16-18\right)i.

$$Re(\frac{-68-34i}{17})$$

Divide -68-34i by 17 to get -4-2i.

$$Re(-4-2i)$$

The real part of -4-2i is -4.

$$-4$$