Solve 2-i/1+i

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

Evaluate

$$\frac{1}{2}-\frac{3}{2}i=0.5-1.5i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 1-3i by 2 to get \frac{1}{2}-\frac{3}{2}i.

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

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 1-3i by 2 to get \frac{1}{2}-\frac{3}{2}i.

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

The real part of \frac{1}{2}-\frac{3}{2}i is \frac{1}{2}.

$$\frac{1}{2}$$