Image of the closed unit disc under a map. What is the image of the closed unit disc i.e., $\{\lambda:|\lambda|\leq1\}$ under the map $\phi(z)=z(z+1)/2$?
 A: The image of the unit circle is the curve with parametric equations $x=\frac12(\cos 2t+\cos t)$,
$y=\frac12(\sin 2t+\sin t)$. This one: 

The inner loop is covered twice. So, the image of the disk is actually this: 

But maybe you are not satisfied with the parametric equation. Let's approach the problem differently: for what values of $w$ does the equation $z^2+z-2w=0$ has a root in $|z|\le 1$? The roots are $-\frac12 \pm \frac12\sqrt{1+8w} $. They are symmetrically located about $-1/2$. Looking at how $-1/2$ sits in the unit disk, we see that the root in which $\sqrt{1+8w}$ has positive real part has a better chance to get in. 
Write $w=x+iy$, $a=1+8x$ and $b=8y$. Wikipedia helpfully gives the formula for the relevant value of $\sqrt{1+8w}$: it is 
$$\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}} + i \operatorname{sign} y \sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}} \tag1$$
Add $-1/2$ to the above, and calculate the absolute value (squared): 
$$\left(-\frac12+\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}\right) + \frac{-a+\sqrt{a^2+b^2}}{2} \tag2$$ 
This has to be $\le 1$. Formula (2) actually simplifies a bit: 
$$\frac14-\frac14\sqrt {2\,a+2\,\sqrt {{a}^{2}+{b}^{2}}}+\frac14\sqrt {{a}^{2}+{
b}^{2}} \tag3$$ 
Plug $a=1+8x$ and $b=8y$ to proudly finish with 
$$
\frac14-\frac14\sqrt {2+16\,x+2\,\sqrt {1+16\,x+64\,{x}^{2}+64\,{y}^{2}}}+ \frac14\sqrt {1+16\,x+64\,{x}^{2}+64\,{y}^{2}}
\le 1 $$
I checked this with an implicit plot function: 

