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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.

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cactus314
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2 Answers2

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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 $\phi=\pi/4$.

orion
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  • what if you cut at positive reals? – cactus314 Jan 29 '16 at 21:29
  • Usually, you don't ;) You probably mean that you interpret the polar angle to lie between [0,2pi) instead of (-pi,pi] (notice how I included $\pi$ in the interval, because we have by almost any definition $\sqrt{-1}=i$ and not $\sqrt{-1}=-i$). So... we want square roots of reals to remain real, as per usual. But anything even slightly below the real line would be interpreted as just below $2\pi$ angle, and would return the negative version of the square root (you're just choosing a different sign at $\pm \sqrt{z}$). – orion Jan 29 '16 at 21:37
  • Anyway, in that case, your identity wouldn't hold, as conjugation wouldn't negate the angle. – orion Jan 29 '16 at 21:37
  • The paper that I am reading is unusual as they are cutting along $\mathbb{R}^+$ instead of $\mathbb{R}^-$ I should have mentioned that thanks. – cactus314 Jan 29 '16 at 21:52
  • Oh, that's the same, just flipped. Just take (-2pi,0]. But I don't get why they don't want $\sqrt{-1}=i$ :) – orion Jan 29 '16 at 21:57
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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:

$$\frac{2}{1+i}=\frac{2(1-i)}{(1+i)(1-i)}=\frac{2-2i}{1^2+1^2}=\frac{2-2i}{2}=\frac{2}{2}-\frac{2i}{2}=\frac{2}{2}-\frac{2}{2}i=1-1i=1-i$$

Jan Eerland
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