Solve 1-2i/2+i=

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

Evaluate

$$-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide -5i by 5 to get -i.

$$-i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide -5i by 5 to get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$