What is this

How do I find out what

$i^{0.5}\\ =(0+1i)^{0.5}\\ =[e^{\frac{\pi}{2}i}]^{0.5}\\ =e^{\frac{\pi}{4}i}\\ =cos\frac{\pi}{4}+isin\frac{\pi}{4}\\ =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\\ =\frac{1+i}{\sqrt{2}}\\$

How do I find out what

$\sqrt {i}$

is?

$\begin{array}{rcll} \sqrt{z} = \sqrt{i} = \ ? \qquad z = i \end{array}$

$\text{Formula}: \begin{array}{|rcll|} \hline z &=& a +i\cdot b & |z| &=& \sqrt{a^2+b^2} & |z| = \text{ magnitude of z } \\ z &=& 0 +i\cdot 1 & |z| &=& \sqrt{0^2+1^2} & \quad a = 0 \quad b= 1\\ && &&=& \sqrt{1} \\ && &&=& 1 \\ \varphi &=& \arctan(\frac{b}{a}) \\ \varphi &=& \arctan(\frac{1}{0}) \quad \Rightarrow \varphi = \frac{\pi}{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline r= \sqrt{z} &=&\sqrt{i} \\ &=& \sqrt{|z|} \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& \sqrt{1} \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& 1 \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& \cos( \frac{\frac{\pi}{2}}{2} ) + i\cdot \sin(\frac{\frac{\pi}{2}}{2}) \\ &=& \cos( \frac{\pi}{4} ) + i\cdot \sin(\frac{\pi}{4}) \\ &=& \frac{\sqrt{2}} {2} + i\cdot \frac{\sqrt{2}} {2} \\\\ \mathbf{r_1=\sqrt{i}} &\mathbf{=}& \mathbf{ \frac{\sqrt{2}} {2} + i\cdot \frac{\sqrt{2}} {2} } \\\\ r_2 &=& -r_1 \\ \mathbf{r_2=\sqrt{i}} &\mathbf{=}& \mathbf{ - \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} } \\\\ \hline \end{array}$

The square root has two solutions!

Here is my formula without compute  angle $\varphi$ :

It doesn't work if $a \ge 0 \land b = 0$ .

Then you can use the normal $\sqrt{a}.$

$\begin{array}{|rcll|} \hline r_1 = \sqrt{z} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{b}{\sqrt{|z|-a}} + i \cdot \sqrt{|z|-a} \right); \qquad r_2 = -r_1 \quad & | \quad z =a+i\cdot b \quad |z| = \sqrt{a^2+b^2}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z &=& i \\ z &=& 0 +i\cdot 1 & |z| = \sqrt{0^2+1^2}= 1 \qquad a = 0 \quad b= 1 \\\\ r_1 &=& \sqrt{i} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{1}{\sqrt{1-0}} + i \cdot \sqrt{1-0} \right) \\ &=& \sqrt{i} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{1}{1} + i \cdot 1 \right) \\ \mathbf{r_1=\sqrt{i}} &\mathbf{=}& \mathbf{ \frac{1}{ \sqrt{2} } \cdot \left( 1 + i \right) } \\\\ r_2 &=& -r_1 \\ \mathbf{r_2=\sqrt{i}} &\mathbf{=}& \mathbf{ -\frac{1}{ \sqrt{2} } \cdot \left( 1 + i \right) } \\ \hline \end{array}$

Because $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ we have the same solutions, see above.

Simplify the following:
sqrt(i)

i = 1/2 + i - 1/2 = (1 + 2 i - 1)/2 = (1 + 2 i + i^2)/2 = (1 + i)^2/2:
 sqrt((1 + i)^2/2 )

sqrt(1/2 (1 + i)^2) = (sqrt((1 + i)^2))/(sqrt(2)):
(sqrt((1 + i)^2))/(sqrt(2))

Cancel exponents. sqrt((1 + i)^2) = 1 + i:
1 + i/sqrt(2)

Rationalize the denominator. (1 + i)/sqrt(2) = (1 + i)/sqrt(2)×(sqrt(2))/(sqrt(2)) = ((1 + i) sqrt(2))/(2):
Answer: |((1 + i) sqrt(2))/(2)