Solve 1-4i/5+12i

$$\frac { 1 - 4 i } { 5 + 12 i }$$

Evaluate

$$-\frac{43}{169}-\frac{32}{169}i\approx -0.25443787-0.189349112i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5-12i.

$$\frac{\left(1-4i\right)\left(5-12i\right)}{\left(5+12i\right)\left(5-12i\right)}$$

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

$$\frac{\left(1-4i\right)\left(5-12i\right)}{5^{2}-12^{2}i^{2}}$$

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

$$\frac{\left(1-4i\right)\left(5-12i\right)}{169}$$

Multiply complex numbers 1-4i and 5-12i like you multiply binomials.

$$\frac{1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)i^{2}}{169}$$

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

$$\frac{1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)\left(-1\right)}{169}$$

Do the multiplications in 1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)\left(-1\right).

$$\frac{5-12i-20i-48}{169}$$

Combine the real and imaginary parts in 5-12i-20i-48.

$$\frac{5-48+\left(-12-20\right)i}{169}$$

Do the additions in 5-48+\left(-12-20\right)i.

$$\frac{-43-32i}{169}$$

Divide -43-32i by 169 to get -\frac{43}{169}-\frac{32}{169}i.

$$-\frac{43}{169}-\frac{32}{169}i$$

Real Part

$$-\frac{43}{169} = -0.25443786982248523$$

Solution Steps

Multiply both numerator and denominator of \frac{1-4i}{5+12i} by the complex conjugate of the denominator, 5-12i.

$$Re(\frac{\left(1-4i\right)\left(5-12i\right)}{\left(5+12i\right)\left(5-12i\right)})$$

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

$$Re(\frac{\left(1-4i\right)\left(5-12i\right)}{5^{2}-12^{2}i^{2}})$$

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

$$Re(\frac{\left(1-4i\right)\left(5-12i\right)}{169})$$

Multiply complex numbers 1-4i and 5-12i like you multiply binomials.

$$Re(\frac{1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)i^{2}}{169})$$

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

$$Re(\frac{1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)\left(-1\right)}{169})$$

Do the multiplications in 1\times 5+1\times \left(-12i\right)-4i\times 5-4\left(-12\right)\left(-1\right).

$$Re(\frac{5-12i-20i-48}{169})$$

Combine the real and imaginary parts in 5-12i-20i-48.

$$Re(\frac{5-48+\left(-12-20\right)i}{169})$$

Do the additions in 5-48+\left(-12-20\right)i.

$$Re(\frac{-43-32i}{169})$$

Divide -43-32i by 169 to get -\frac{43}{169}-\frac{32}{169}i.

$$Re(-\frac{43}{169}-\frac{32}{169}i)$$

The real part of -\frac{43}{169}-\frac{32}{169}i is -\frac{43}{169}.

$$-\frac{43}{169}$$