Solve 4-i/2+i

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

Evaluate

$$\frac{7}{5}-\frac{6}{5}i=1.4-1.2i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-i.

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

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

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

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

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

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

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

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

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

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

$$\frac{8-4i-2i-1}{5}$$

Combine the real and imaginary parts in 8-4i-2i-1.

$$\frac{8-1+\left(-4-2\right)i}{5}$$

Do the additions in 8-1+\left(-4-2\right)i.

$$\frac{7-6i}{5}$$

Divide 7-6i by 5 to get \frac{7}{5}-\frac{6}{5}i.

$$\frac{7}{5}-\frac{6}{5}i$$

Real Part

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

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{8-4i-2i-1}{5})$$

Combine the real and imaginary parts in 8-4i-2i-1.

$$Re(\frac{8-1+\left(-4-2\right)i}{5})$$

Do the additions in 8-1+\left(-4-2\right)i.

$$Re(\frac{7-6i}{5})$$

Divide 7-6i by 5 to get \frac{7}{5}-\frac{6}{5}i.

$$Re(\frac{7}{5}-\frac{6}{5}i)$$

The real part of \frac{7}{5}-\frac{6}{5}i is \frac{7}{5}.

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