Solve 21-i/2+i=

$$2 \frac { 1 - i } { 2 + i } =$$

Evaluate

$$\frac{2}{5}-\frac{6}{5}i=0.4-1.2i$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{2+i} by the complex conjugate of the denominator, 2-i.

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

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

$$2\times \frac{\left(1-i\right)\left(2-i\right)}{2^{2}-i^{2}}$$

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

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

Multiply complex numbers 1-i and 2-i like you multiply binomials.

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

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

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

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

$$2\times \frac{2-i-2i-1}{5}$$

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

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

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

$$2\times \frac{1-3i}{5}$$

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

$$2\left(\frac{1}{5}-\frac{3}{5}i\right)$$

Multiply 2 times \frac{1}{5}-\frac{3}{5}i.

$$2\times \frac{1}{5}+2\times \left(-\frac{3}{5}i\right)$$

Do the multiplications.

$$\frac{2}{5}-\frac{6}{5}i$$

Real Part

$$\frac{2}{5} = 0.4$$

Solution Steps

Multiply both numerator and denominator of \frac{1-i}{2+i} by the complex conjugate of the denominator, 2-i.

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

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

$$Re(2\times \frac{\left(1-i\right)\left(2-i\right)}{2^{2}-i^{2}})$$

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

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

Multiply complex numbers 1-i and 2-i like you multiply binomials.

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

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

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

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

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

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

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

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

$$Re(2\times \frac{1-3i}{5})$$

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

$$Re(2\left(\frac{1}{5}-\frac{3}{5}i\right))$$

Multiply 2 times \frac{1}{5}-\frac{3}{5}i.

$$Re(2\times \frac{1}{5}+2\times \left(-\frac{3}{5}i\right))$$

Do the multiplications in 2\times \frac{1}{5}+2\times \left(-\frac{3}{5}i\right).

$$Re(\frac{2}{5}-\frac{6}{5}i)$$

The real part of \frac{2}{5}-\frac{6}{5}i is \frac{2}{5}.

$$\frac{2}{5}$$