Solve 3-i/3+4i

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

Evaluate

$$\frac{1}{5}-\frac{3}{5}i=0.2-0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-4i.

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

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

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

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

$$\frac{\left(3-i\right)\left(3-4i\right)}{25}$$

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

$$\frac{3\times 3+3\times \left(-4i\right)-i\times 3-\left(-4i^{2}\right)}{25}$$

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

$$\frac{3\times 3+3\times \left(-4i\right)-i\times 3-\left(-4\left(-1\right)\right)}{25}$$

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

$$\frac{9-12i-3i-4}{25}$$

Combine the real and imaginary parts in 9-12i-3i-4.

$$\frac{9-4+\left(-12-3\right)i}{25}$$

Do the additions in 9-4+\left(-12-3\right)i.

$$\frac{5-15i}{25}$$

Divide 5-15i by 25 to get \frac{1}{5}-\frac{3}{5}i.

$$\frac{1}{5}-\frac{3}{5}i$$

Real Part

$$\frac{1}{5} = 0.2$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(3-i\right)\left(3-4i\right)}{25})$$

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

$$Re(\frac{3\times 3+3\times \left(-4i\right)-i\times 3-\left(-4i^{2}\right)}{25})$$

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

$$Re(\frac{3\times 3+3\times \left(-4i\right)-i\times 3-\left(-4\left(-1\right)\right)}{25})$$

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

$$Re(\frac{9-12i-3i-4}{25})$$

Combine the real and imaginary parts in 9-12i-3i-4.

$$Re(\frac{9-4+\left(-12-3\right)i}{25})$$

Do the additions in 9-4+\left(-12-3\right)i.

$$Re(\frac{5-15i}{25})$$

Divide 5-15i by 25 to get \frac{1}{5}-\frac{3}{5}i.

$$Re(\frac{1}{5}-\frac{3}{5}i)$$

The real part of \frac{1}{5}-\frac{3}{5}i is \frac{1}{5}.

$$\frac{1}{5}$$