sqrt(1+i) Is this proof correct?

Complex root:

\(\begin{array}{|rcll|} \hline x &=&\sqrt{1+i}\\ \ln x &=& \dfrac{1}{2}\cdot \ln(1+i)\\ 2\cdot \ln x &=& \ln(1+i)\\ &=& \ln(|1+i|)+i\arg(1+i) \quad & | \quad |1+i| = \sqrt{1^2+1^2}= \sqrt{2} \\ &=& \ln(\sqrt{2})+i\cdot \arg(1+i) \quad & | \quad \arg(1+i) = \arctan(\frac{1}{1}) = \dfrac{\pi}{4} \\ &=& \ln(\sqrt{2}) + i\cdot \dfrac{\pi}{4} \\ 2\cdot \ln x &=&\ln(\sqrt{2}) + i\cdot \dfrac{\pi}{4} \quad & | \quad :2 \\ \ln x &=& \dfrac12 \cdot \ln(\sqrt2) + i\frac{\pi}{8} \\ &=& \ln\left(\sqrt{\sqrt{2}}\right) + i \frac{\pi}{8} \\ \ln x &=& \ln(\sqrt[4]{2}) + i \frac{\pi}{8} \\ x &=&e^{\ln(\sqrt[4]{2}) + i \frac{\pi}{8}} \\ &=&e^{\ln(\sqrt[4]{2})} e^{i \frac{\pi}{8}} \\ x &=& \sqrt[4]{2} \cdot e^{i \frac{\pi}{8}} \\ x_1 &=& \sqrt[4]{2} \cdot (\cos{\frac{\pi}{8}} +i \cdot \sin{\frac{\pi}{8}} )\\ && \cos{\frac{\pi}{8}} = \frac12\cdot \sqrt{2+\sqrt{2}} \\ && \sin{\frac{\pi}{8}} = \frac12\cdot \sqrt{2-\sqrt{2}} \\ x_1 &=& \sqrt[4]{2} \cdot \left( \frac12\cdot \sqrt{2+\sqrt{2}} +i \cdot \frac12\cdot \sqrt{2-\sqrt{2}} \right) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2} +i \cdot \sqrt[4]{2}\cdot \frac{\sqrt{2-\sqrt{2}}}{2} } \\\\ x_2 &=& - x_1 \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2} - i \cdot \sqrt[4]{2}\cdot \frac{\sqrt{2-\sqrt{2}}}{2} } \\ \hline \end{array}\)

\(x_1 \approx 1.09868 + 0.45509i \\ x_2 \approx -1.09868 - 0.45509i \)

Easier to convert to polar form and then use de Moivre's theorem, (cos x   +  i sin x)^n

= (cos nx  + i sin nx)

(1 + i )   = sqrt2( cos pi/4   + isin pi/4)

(1 + i)^1/2   = {(sqrt2)^1/2 }(cos pi/8  +i sin pi/8)