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

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

Evaluate

$$-\frac{3}{5}+\frac{1}{5}i=-0.6+0.2i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Multiply i times 1+i.

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

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

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

Reorder the terms.

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

Divide 1-i by -1+i to get -1.

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

Subtract 1 from \frac{2}{5}+\frac{1}{5}i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from \frac{2}{5} to get -\frac{3}{5}.

$$-\frac{3}{5}+\frac{1}{5}i$$

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.

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

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

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

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

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

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

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

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

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

Multiply i times 1+i.

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

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

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

Reorder the terms.

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

Divide 1-i by -1+i to get -1.

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

Subtract 1 from \frac{2}{5}+\frac{1}{5}i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from \frac{2}{5} to get -\frac{3}{5}.

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

The real part of -\frac{3}{5}+\frac{1}{5}i is -\frac{3}{5}.

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