Square root of square root of -1

Method 1: Since $i = e^{i*\frac{\pi}{2}}$ we have that

$\sqrt{i} = e^{i*\frac{\pi}{4}} = \cos(\frac{\pi}{4}) + i*\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + i*\frac{\sqrt{2}}{2} = a + bi$ ,

and so $(a + b)^{2} = (2*\frac{\sqrt{2}}{2})^{2} = \boxed{2}$ .

Method 2: With $\sqrt{i} = a + bi$ , square both sides of this equation to find that

$i = a^{2} - b^{2} + 2abi$ .

Next, equating coefficients of real and complex terms from each side, we have that

$a^{2} - b^{2} = 0 \Longrightarrow a^{2} = b^{2}$ and $2ab = 1$ .

The second of these equations tells us that $a$ and $b$ must be of the same sign, so from the first equation we can conclude that $a = b$ . Thus $2a^{2} = 1$ , and so $(a + b)^{2} = (2a)^{2} = 4a^{2} = \boxed{2}$ .

$\begin{array}{l} {\sqrt{i}\mathrm{{=}}{a}\mathrm{{+}}{bi}\hspace{0.33em}\hspace{0.33em}{\mathrm{,}}\hspace{0.33em}}\\ {{i}\hspace{0.33em}\mathrm{{=}}{a}^{2}\mathrm{{-}}{b}^{2}\mathrm{{+}}{2}{abi}}\\ {{2}{abi}\mathrm{{=}}{i}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{and}\hspace{0.33em}\hspace{0.33em}{a}^{2}\mathrm{{=}}{b}^{2}}\\ {{therefore}\hspace{0.33em}{2}{ab}\mathrm{{=}}{1}\hspace{0.33em}{and}\hspace{0.33em}{b}\mathrm{{=}}\mathrm{\pm}{a}}\\ {{2}{a}\left({\mathrm{\pm}{a}}\right)\mathrm{{=}}{1}}\\ {{therefore}\hspace{0.33em}{a}\mathrm{{=}}\mathrm{\pm}\frac{1}{\sqrt{2}}\mathrm{{=}}{b}\hspace{0.33em}}\\ {{therefore}\hspace{0.33em}{\mathrm{(}}{a}\mathrm{{+}}{b}{\mathrm{)}}^{2}\mathrm{{=}}\hspace{0.33em}{\mathrm{(}}\frac{1}{\sqrt{2}}\mathrm{{+}}\frac{1}{\sqrt{2}}{\mathrm{)}}^{2}\mathrm{{=}}{\mathrm{(}}\frac{2}{\sqrt{2}}{\mathrm{)}}^{2}\mathrm{{=}}{\mathrm{(}}\frac{4}{2}{\mathrm{)}}\mathrm{{=}}\fbox{\${2}\$}} \end{array}$

Assume that $\displaystyle \sqrt{i} = x+iy$ for some $x,y \in \Re$ .

Then $\displaystyle i = x^2 - y^2 + 2ixy$

Comparing real and imaginary parts, we get:

$\displaystyle x^2 - y^2 = 1; 2xy = 1$

Using the identity $\displaystyle (x^2-y^2) = \sqrt{(x^2+y^2)^2 - (2xy)^2}$ , we get:

$\displaystyle (x^2-y^2) = 0 \Rightarrow x=\frac{1}{\sqrt{2}}=y$ or $\displaystyle x=-\frac{1}{\sqrt{2}}=y$ .

Here, both must be of the same sign since $2xy>0$ .

Hence, $\displaystyle \sqrt{i} = \pm\frac{1}{\sqrt{2}} \pm\frac{1}{\sqrt{2}}i$

Finally giving the answer $(a+b)^2= (\pm\sqrt{2})^2 = \boxed{2}$

Its simple square the relation. I=a²-b²+2abi then we get comparing coefficients a=b a=-b not possible. 2ab=1 a=b=±1/(2)½

Here's a pretty simple way I did in my head:

With complex numbers, to find the product of two of them you multiply their norms and add the angles to the x-axis. So, you want the complex number with norm 1 that is at 45 degrees to the x-axis, which happens to make a very familiar triangle. The hypotenuse is 1, so each side is $\frac { \sqrt { 2 } }{ 2 }$ . Therefore, $a+b=\sqrt { 2 }$ , so the square of $a+b$ is $2$ .