Solve (1-2i)-1/2+i

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

Evaluate

$$\frac{3}{5}-\frac{9}{5}i=0.6-1.8i$$

Solution Steps

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

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

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

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

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

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

Multiply 1 and 2-i to get 2-i.

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

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

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

Combine the real and imaginary parts in numbers 1-2i and -\frac{2}{5}+\frac{1}{5}i.

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

Add 1 to -\frac{2}{5}. Add -2 to \frac{1}{5}.

$$\frac{3}{5}-\frac{9}{5}i$$

Real Part

$$\frac{3}{5} = 0.6$$

Solution Steps

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

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

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

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

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

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

Multiply 1 and 2-i to get 2-i.

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

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

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

Combine the real and imaginary parts in numbers 1-2i and -\frac{2}{5}+\frac{1}{5}i.

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

Add 1 to -\frac{2}{5}. Add -2 to \frac{1}{5}.

$$Re(\frac{3}{5}-\frac{9}{5}i)$$

The real part of \frac{3}{5}-\frac{9}{5}i is \frac{3}{5}.

$$\frac{3}{5}$$