Does $\overline{ \sqrt{1 + i}} = \sqrt{1-i} \ $? I am having trouble with complex conjugates today.  Can someone help me?  $$\overline{ \sqrt{1 + i}} \stackrel{\color{#2222FF}{?}}{=} \sqrt{1-i} \tag{$\ast$} $$
In this case, since $\cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}(i+i)$.  Then we can take the square root and get:
$$\cos \frac{\pi}{8} + i \sin \frac{\pi}{8} = \frac{1}{\sqrt[4]{2}}\sqrt{1+i}$$
Then if we take complex conjugate of both sides, I don't know the cosine of $\frac{\pi}{8} = 22.5^\circ$ off the top of my head
$$\cos \frac{\pi}{8} - i \sin \frac{\pi}{8} = \frac{1}{\sqrt[4]{2}} \overline{\sqrt{1+i}}$$
On other parts of Math.SE (and this) we find that $\cos \frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ a very interesting number indeed.
$$\frac{1}{\sqrt[4]{2}}\sqrt{1+i} =  \frac{\sqrt{2 + \sqrt{2}}}{2} + i \frac{\sqrt{2 - \sqrt{2}}}{2}$$
At this point, I am still not sure $(\ast)$ is correct or not.

Notice, if $z=a+bi$ and $a,b\in\mathbb{R}$:


*

*$$\overline{z}z=\Re^2(z)+\Im^2(z)$$

*$$\overline{z}=\frac{|z|^2}{z}$$

*$$|z|=\sqrt{\Re^2(z)+\Im^2(z)}$$



$$\overline{\sqrt{1+i}}=\frac{|\sqrt{1+i}|^2}{\sqrt{1+i}}=\frac{|\left(1+i\right)^{\frac{1}{2}}|^2}{\sqrt{1+i}}=\frac{|1+i|}{\sqrt{1+i}}=\frac{\sqrt{1^2+1^2}}{\sqrt{1+i}}=\frac{\sqrt{2}}{\sqrt{1+i}}=\sqrt{\frac{2}{1+i}}=\sqrt{1-i}$$
Because:
$$\frac{2}{1+i}=\frac{2(1-i)}{(1+i)(1-i)}=\frac{2-2i}{1^2+1^2}=\frac{2-2i}{2}=\frac{2}{2}-\frac{2i}{2}=\frac{2}{2}-\frac{2}{2}i=1-1i=1-i$$
