Solve 13+3i/12+5i=

$$\frac { 13 + 3 i } { 12 + 5 i } =$$

Evaluate

$$\frac{171}{169}-\frac{29}{169}i\approx 1.01183432-0.171597633i$$

Solution Steps

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

$$\frac{\left(13+3i\right)\left(12-5i\right)}{\left(12+5i\right)\left(12-5i\right)}$$

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

$$\frac{\left(13+3i\right)\left(12-5i\right)}{12^{2}-5^{2}i^{2}}$$

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

$$\frac{\left(13+3i\right)\left(12-5i\right)}{169}$$

Multiply complex numbers 13+3i and 12-5i like you multiply binomials.

$$\frac{13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)i^{2}}{169}$$

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

$$\frac{13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)\left(-1\right)}{169}$$

Do the multiplications in 13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)\left(-1\right).

$$\frac{156-65i+36i+15}{169}$$

Combine the real and imaginary parts in 156-65i+36i+15.

$$\frac{156+15+\left(-65+36\right)i}{169}$$

Do the additions in 156+15+\left(-65+36\right)i.

$$\frac{171-29i}{169}$$

Divide 171-29i by 169 to get \frac{171}{169}-\frac{29}{169}i.

$$\frac{171}{169}-\frac{29}{169}i$$

Real Part

$$\frac{171}{169} = 1\frac{2}{169} = 1.0118343195266273$$

Solution Steps

Multiply both numerator and denominator of \frac{13+3i}{12+5i} by the complex conjugate of the denominator, 12-5i.

$$Re(\frac{\left(13+3i\right)\left(12-5i\right)}{\left(12+5i\right)\left(12-5i\right)})$$

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

$$Re(\frac{\left(13+3i\right)\left(12-5i\right)}{12^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{\left(13+3i\right)\left(12-5i\right)}{169})$$

Multiply complex numbers 13+3i and 12-5i like you multiply binomials.

$$Re(\frac{13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)i^{2}}{169})$$

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

$$Re(\frac{13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)\left(-1\right)}{169})$$

Do the multiplications in 13\times 12+13\times \left(-5i\right)+3i\times 12+3\left(-5\right)\left(-1\right).

$$Re(\frac{156-65i+36i+15}{169})$$

Combine the real and imaginary parts in 156-65i+36i+15.

$$Re(\frac{156+15+\left(-65+36\right)i}{169})$$

Do the additions in 156+15+\left(-65+36\right)i.

$$Re(\frac{171-29i}{169})$$

Divide 171-29i by 169 to get \frac{171}{169}-\frac{29}{169}i.

$$Re(\frac{171}{169}-\frac{29}{169}i)$$

The real part of \frac{171}{169}-\frac{29}{169}i is \frac{171}{169}.

$$\frac{171}{169}$$