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

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

Evaluate

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

Solution Steps

Multiply 1-2i times i.

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

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

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

Do the multiplications in i-2\left(-1\right). Reorder the terms.

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

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.

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

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

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

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

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

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

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

Real Part

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Multiply 1-2i times i.

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

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

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

Do the multiplications in i-2\left(-1\right). Reorder the terms.

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

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.

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

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

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

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

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

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

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

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

$$\frac{3}{2}$$