Answer
$1+i~~~$ is not in the Mandelbrot set.
Work Step by Step
$z = 1+i$
We can find the value of $z^2+z$:
$z^2+z = (1+i)^2+(1+i)$
$z^2+z = (1+2i-1)+(1+i)$
$z^2+z = 1+3i$
We can find the absolute value of $1+3i$:
$\sqrt{1^2+3^2} = \sqrt{10}$
Note that $\sqrt{10} \gt 2$.
Therefore $~~1+i~~$ is not in the Mandelbrot set.