Solve 1+2i/1+i+2iz

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

Evaluate

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

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{3+i}{2}+2iz$$

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

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

Expand

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

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{3+i}{2}+2iz$$

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

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