(a+bi)^2=i What is the squareroot of i solve geometrically in a complex number plane
One method:
$\displaystyle \displaystyle \begin{align*} (a + b\,i)^2 &= i \\ a^2 - b^2 + 2ab\,i &= 0 + 1i \\ a^2 - b^2 = 0 \textrm{ and } 2ab &= 1 \end{align*}$
From the second of these equations we have $\displaystyle \displaystyle \begin{align*} b = \frac{1}{2a} \end{align*}$. Substituting into the first equation we have
$\displaystyle \displaystyle \begin{align*} a^2 - \left(\frac{1}{2a}\right)^2 &= 0 \\ a^2 - \frac{1}{4a^2} &= 0 \\ a^2 &= \frac{1}{4a^2} \\ 4a^4 &= 1 \\ a^4 &= \frac{1}{4} \\ a &= \pm \frac{\sqrt{2}}{2} \end{align*}$
Back-substituting gives
$\displaystyle \displaystyle \begin{align*} b &= \frac{1}{2\left( \pm \frac{\sqrt{2}}{2}\right) } \\ &= \frac{1}{ \pm \sqrt{2}} \\ &= \pm \frac{\sqrt{2}}{2} \end{align*}$
Therefore the solutions to your equation are $\displaystyle \displaystyle \begin{align*} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \end{align*}$ and $\displaystyle \displaystyle \begin{align*} -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \end{align*}$.
These are consistent with the solutions given by Plato, who probably used the Exponential form of each number and Index Laws as an alternative method.
I don't understand what the problem is. You will always need to use some algebra to at least find one solution. If you wish to use more geometry after that, you can use the fact that there are as many "roots" as its power (i.e. square roots have 2 solutions, cube roots have 3 solutions, fourth roots have 4 solutions, etc) and they are all evenly spaced around a circle centred at the origin.