Solve 2*1+i/2+2*i

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

Evaluate

$$1$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$2\times \frac{4}{8}$$

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

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

Cancel out 2 and 2.

$$1$$

Real Part

$$1$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$Re(2\times \frac{4}{8})$$

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

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

Cancel out 2 and 2.

$$Re(1)$$

The real part of 1 is 1.

$$1$$