Solve 1+i/2+2i

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

Evaluate

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

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-2i.

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

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

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

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

$$\frac{\left(1+i\right)\left(2-2i\right)}{8}$$

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

$$\frac{1\times 2+1\times \left(-2i\right)+2i-2i^{2}}{8}$$

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

$$\frac{1\times 2+1\times \left(-2i\right)+2i-2\left(-1\right)}{8}$$

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

$$\frac{2-2i+2i+2}{8}$$

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

$$\frac{2+2+\left(-2+2\right)i}{8}$$

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

$$\frac{4}{8}$$

Divide 4 by 8 to get \frac{1}{2}.

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

Real Part

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

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(1+i\right)\left(2-2i\right)}{8})$$

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

$$Re(\frac{1\times 2+1\times \left(-2i\right)+2i-2i^{2}}{8})$$

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

$$Re(\frac{1\times 2+1\times \left(-2i\right)+2i-2\left(-1\right)}{8})$$

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

$$Re(\frac{2-2i+2i+2}{8})$$

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

$$Re(\frac{2+2+\left(-2+2\right)i}{8})$$

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

$$Re(\frac{4}{8})$$

Divide 4 by 8 to get \frac{1}{2}.

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

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

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