Complex Number Arithmetic

Problem 182: Show that $(1+z)^2 = 1 + 2z + z^2$ where $z =(x+iy)$ .

Solution: Using the properties of complex numbers we will solve our problem. That is, if $z_1=x_1+iy_1$ and $z_2 = x_2 + i y_2$ are two complex numbers, then

(1) $\begin{equation*}z_1 \cdot z_2 = (x_1 + iy_1) \cdot (x_2 + iy_2) = (x_1x_2 - y_1y_2) + i(x_2y_1 + y_2x_1). \end{equation*}$

Likewise,

(2) $\begin{align*} (1+z)^2 & = (1+z)(1+z) \\ & = \left[1 +(x+iy) \right] \cdot \left[1 +(x+iy) \right] \\ & = \left[ (1+x) + iy \right] \cdot \left[ (1 +x) +iy) \right] \\ & = \left[ (1+x)(1+x) - y^2 \right] + i \left[ y(1+x) + y(1+x) \right] \\ & = 1 + 2x + x^2 - y^2 + i \left( y + yx + y + yx \right) \\ & = 1 + 2x + x^2 - y^2 + i(2y + 2yx) \\ & = 1 + 2x + 2y i + 2yxi - y^2 + x^2 \\ & = 1 + 2\underbrace{(x+iy)}_{= z} + \underbrace{x^2-y^2+2yxi}_{= z^2} \\ & = 1 +2z + z^2.\end{align*}$

That is,

(3) $\begin{equation*} (1+z)^2 = 1 + 2z + z^2 \end{equation*}$

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