Nested Radicals (4)

This is just the solution to the bonus. I leave the main solution up to the solvers.

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First, we must define $\sqrt{-.25}$ in terms of $i$ , which is $\sqrt{\frac{1}{4}}\sqrt{-1}=.5i$ . Next, we define $\sqrt{-.25+.5i}$ as $a+bi$ , where $a$ and $b$ are real numbers: $\sqrt{-.25+.5i}=a+bi$ $-.25+.5i=a^2-b^2+2abi$

It might not be obvious, but we now have a system of equations. If we separate it into the real and imaginary parts, $a^2-b^2=-.25$ $.5i=2abi$ First, start with the imaginary equation: $.5i=2abi$ $b=\frac{1}{4a}$

By substitution into the real equation, $a^2-\left(\frac{1}{4a}\right)^2=-.25$

Simplify, multiply both sides by $a^2$ , and set $a^2=c$ : $c^2+\frac{1}{4}c-\frac{1}{16}=0$

Apply the quadratic formula and heavily simplify, $c=a^2=\frac{\sqrt{5}-1}{8}$ $a=\sqrt{\frac{\sqrt{5}-1}{8}}$

We're nearly there! Using the real equation, we can find $b$ : $b^2=-\left(\frac{-1}{4}-\frac{\sqrt{5}-1}{8}\right)$ $b=\sqrt{\frac{\sqrt{5}+1}{8}}$

Finally! Putting this all together, we get: $\sqrt{-.25+\sqrt{-.25}} = \boxed{\sqrt{\frac{\sqrt{5}-1}{8}}+\sqrt{\frac{\sqrt{5}+1}{8}}i}$

We can check this with the WolframAlpha link in the problem, too.