Solve 4-2i/3-i

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

Evaluate

$$\frac{7}{5}-\frac{1}{5}i=1.4-0.2i$$

Solution Steps

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

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

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

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

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

$$\frac{\left(4-2i\right)\left(3+i\right)}{10}$$

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

$$\frac{4\times 3+4i-2i\times 3-2i^{2}}{10}$$

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

$$\frac{4\times 3+4i-2i\times 3-2\left(-1\right)}{10}$$

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

$$\frac{12+4i-6i+2}{10}$$

Combine the real and imaginary parts in 12+4i-6i+2.

$$\frac{12+2+\left(4-6\right)i}{10}$$

Do the additions in 12+2+\left(4-6\right)i.

$$\frac{14-2i}{10}$$

Divide 14-2i by 10 to get \frac{7}{5}-\frac{1}{5}i.

$$\frac{7}{5}-\frac{1}{5}i$$

Real Part

$$\frac{7}{5} = 1\frac{2}{5} = 1.4$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(4-2i\right)\left(3+i\right)}{10})$$

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

$$Re(\frac{4\times 3+4i-2i\times 3-2i^{2}}{10})$$

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

$$Re(\frac{4\times 3+4i-2i\times 3-2\left(-1\right)}{10})$$

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

$$Re(\frac{12+4i-6i+2}{10})$$

Combine the real and imaginary parts in 12+4i-6i+2.

$$Re(\frac{12+2+\left(4-6\right)i}{10})$$

Do the additions in 12+2+\left(4-6\right)i.

$$Re(\frac{14-2i}{10})$$

Divide 14-2i by 10 to get \frac{7}{5}-\frac{1}{5}i.

$$Re(\frac{7}{5}-\frac{1}{5}i)$$

The real part of \frac{7}{5}-\frac{1}{5}i is \frac{7}{5}.

$$\frac{7}{5}$$