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.
 A: How much exactly that cosine is, is beside the point. Two things are important here. One is the fact, that you have to know which square root you take (in complex, we usually cut at the negative reals, so square roots of numbers in I and II quadrant lie in I quadrant, and square roots of numbers in III and IV quadrants lie in IV quadrant. With that settled, just observe a general complex number in the exponential form:
$$z=|z|e^{i\phi}$$
Now, observe:
$$\sqrt{\overline{z}}=\sqrt{|z|e^{-i\phi}}=\sqrt{|z|} e^{-i\phi/2}$$
and
$$\overline{\sqrt{z}}=\overline{\sqrt{|z|e^{i\phi}}}=\sqrt{|z|} e^{-i\phi/2}$$
So yes... it's the same, as long as $\phi\in (-\pi,\pi)$. In your case, $|z|=\sqrt2$ and $\phi=\pi/4$.
A: 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:
$$\frac{2}{1+i}=\frac{2(1-i)}{(1+i)(1-i)}=\frac{2-2i}{1^2+1^2}=\frac{2-2i}{2}=\frac{2}{2}-\frac{2i}{2}=\frac{2}{2}-\frac{2}{2}i=1-1i=1-i$$
