Solve frac{(1-texttt{i})(1+2texttt{i})}{(1+texttt{i})}

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

Evaluate

$$2-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply complex numbers 3+i and 1-i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\left(-i\right)+i-\left(-1\right).

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

Combine the real and imaginary parts in 3-3i+i+1.

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

Do the additions in 3+1+\left(-3+1\right)i.

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

Divide 4-2i by 2 to get 2-i.

$$2-i$$

Real Part

$$2$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Multiply both numerator and denominator of \frac{3+i}{1+i} by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

Multiply complex numbers 3+i and 1-i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\left(-i\right)+i-\left(-1\right).

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

Combine the real and imaginary parts in 3-3i+i+1.

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

Do the additions in 3+1+\left(-3+1\right)i.

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

Divide 4-2i by 2 to get 2-i.

$$Re(2-i)$$

The real part of 2-i is 2.

$$2$$