Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$

Question

How can I simplify $\sqrt{1+\sqrt{i}}$?

I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$

But I'm not really sure

Answer 1

Using polar coordinates $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt{1+\sqrt{e^{i \pi/2}}} = \sqrt{1 + e^{i \pi/4}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \, $ from this we have

$\displaystyle r^2 = \left( 1 + \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 = 1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2} = 2 + \sqrt{2} \implies r = \sqrt{2 + \sqrt{2}} $

and

$\displaystyle \tan \theta = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 \implies \theta = \arctan(\sqrt{2} - 1)$

So, $\displaystyle \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} = \sqrt{re^{i\theta}} = \sqrt{r} \, e^{i\theta/2}$

Finally, $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt[4]{2 + \sqrt{2}} \, \left( \cos\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) + i\sin\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) \right)$

We can also obtain cartesian form:

$\displaystyle a + ib = \sqrt{1+\sqrt{i}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}}$ yields

$\displaystyle a^2 + b^2 +i(-2ab) = 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \implies a^2 + b^2 = 1 + \frac{\sqrt{2}}{2}$ and $\displaystyle -2ab = \frac{\sqrt{2}}{2}$. So we have

$\displaystyle (a-b)^2 = 1 + \sqrt{2} \,$ and $\displaystyle (a+b)^2 = 1 $ which gives $\displaystyle a = \frac{1 + \sqrt{1 + \sqrt{2}}}{2} $ and $\displaystyle b = \frac{-1 + \sqrt{1 + \sqrt{2}}}{2}$

Answer 2

$$(1+i^\frac{1}{2})^\frac{1}{2}=(1+(e^{i\frac\pi2+i2\pi n})^\frac12 )^\frac12 =(1+e^{i\frac\pi4+i\pi n} )^\frac12$$

$$=\left(1+\cos(\frac\pi4 +\pi n)+ i \sin(\frac\pi4 +\pi n) \right)^\frac12 =(re^{i\theta})^\frac12\tag 1$$

where, $$r=\sqrt{\left(1+\cos(\frac\pi4 +\pi n)\right)^2+\sin^2(\frac\pi4 +\pi n) } =\sqrt{2+2\cos(\frac\pi4 +\pi n) }=2| \cos(\frac\pi8 +\frac{\pi n}2) |$$

$$ \tan\theta = \frac{\sin(\frac\pi4 +\pi n)}{1+\cos(\frac\pi4 +\pi n)} = \tan (\frac\pi8 +\frac{\pi n}2)\implies \theta =\frac\pi8 +\frac{\pi n}2+k\pi $$

Substitute $r$ and $\theta$ into (1),

$$(1+i^\frac{1}{2})^\frac{1}{2} =r^\frac12 e^{i\frac{\theta}2} = \sqrt{2| \cos(\frac\pi8 +\frac{\pi n}2) |} \>e^{i\pi(\frac1{16} +\frac{ n}4+\frac{k}2)}$$

which assumes multiple values. For the special case $n=k=0$,

$$(1+i^\frac{1}{2})^\frac{1}{2} = \sqrt{2\cos\frac\pi8}\>e^{\frac{i\pi}{16}} =\sqrt{2\cos\frac\pi8}\>(\cos\frac{\pi}{16} + i\sin\frac{\pi}{16})$$

Answer 3

I personally think that to see what’s really going on, you do not want to use the polar representation, but rather take your own original strategy to its logical end. If you want quantitative results, the polar method is fine, but pure thought gives you qualitative information that’s just as illuminating.

Thus I would start with the innermost quantity, which you have designated $1+\sqrt i$, and call it $w$, so that $(w-1)^2=i$; then your final quantity is $\sqrt w$, so that if you call this $x$, you have $x^2=w$. Putting it all together, you have $(x^2-1)^2=i$, and we may rewrite this $$ x^4-2x+1-i=0\,, $$ a quartic polynomial over the Gaussian integers $\Bbb Z[i]$, and we know it’s irreducible there by Eisenstein. This means that all four roots are in some sense equally good (or bad).

If you want to have a feel for where they lie in the Wessel-Argand-Gauss plane, start with the square roots of $i$, which are $\pm\frac{1+i}{\sqrt2}$, at unit distance from the origin. Adding $1$ to the root in the third quadrant gets you into the fourth quadrant, but inside the unit circle, and taking the square root of that gets you numbers in fourth and second quadrants, still inside unit circle. If you start with the $\sqrt i$ in the first quadrant, adding $1$ gets you outside the unit circle, and the two square roots are outside the unit circle, in quadrants I and III.