# Question via email about complex numbers

#### Prove It

##### Well-known member
MHB Math Helper
Plot the image of the region \displaystyle \begin{align*} \left| z \right| \geq 5 \end{align*} under the mapping \displaystyle \begin{align*} w = z^2 \end{align*}.
We should note that we can write any complex number as \displaystyle \begin{align*} z = r\,\mathrm{e}^{\mathrm{i}\,\theta} \end{align*} where \displaystyle \begin{align*} r = \left| z \right| \end{align*} and \displaystyle \begin{align*} \theta = \textrm{arg}\,\left( z \right) + 2\,\pi\,n , \,\, n \in \mathbf{Z} \end{align*}. So that means

\displaystyle \begin{align*} z &= r\,\mathrm{e}^{\mathrm{i}\,\theta} \\ \\ z^2 &= \left( r\, \mathrm{e}^{\mathrm{i}\,\theta} \right) ^2 \\ &= r^2\,\mathrm{e}^{2\,\mathrm{i}\,\theta} \end{align*}

thus if \displaystyle \begin{align*} \left| z \right| = r \geq 5 \end{align*} then that means \displaystyle \begin{align*} \left| z^2 \right| = r^2 \geq 25 \end{align*}. Since \displaystyle \begin{align*} \theta \end{align*} can take on any value, that means that \displaystyle \begin{align*} 2\,\theta \end{align*} also can, and thus the region defined by \displaystyle \begin{align*} w = z^2 \end{align*} must be everything on or outside of the circle defined by \displaystyle \begin{align*} \left| z^2 \right| \geq 25 \end{align*}, so in other words, everything on or outside the circle centred at the origin of radius 25 units.

So graphically we have (using the standard convention of shading the region that is not required)... 