Solve 2-i/i(1+i)

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

Evaluate

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

Solution Steps

Multiply i times 1+i.

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

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

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

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}.

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

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.

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

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

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

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

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

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 i times 1+i.

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

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

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

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}.

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

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.

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

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

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

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

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

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}.

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