I know that the square root of complex number is a multivalued function, and by definition: $\sqrt{z}=w\iff w^2=z$ i.e. the square root of z are the solutions of complex equation $w^2=z$.


I would like to evaluate the argument(s) of


using the properties of the principal value of $\arg(\cdot)\in (-\pi, \pi]$.

By definition $$w=\sqrt{1+i}\implies w^2=1+i$$ so $\arg(w^2)=\arg(1+i)$. It's easy to show that $\arg(1+i)=\frac{\pi}{4}$ hence $$\arg(w^2)=\arg(1+i)\iff \arg(w^2)=\frac{\pi}{4}$$

Now I use the property $\arg(w^n)=n\arg(w)$ so the equation becomes $$2\arg(w)=\frac{\pi}{4}\implies \arg(w)=\frac{\pi}{8}.$$ The problem here is that $\frac{\pi}{8}$ is the argument of only one square root of $\sqrt{1+i}$. I lost the argument of the other one... but where?

  • $\begingroup$ Not sure I agree $\sqrt{z}=w\iff w^2=z$. As far as I know, $\sqrt{}$ is never multivalued. Else it wouldn't be a function. $\endgroup$ – Shuri2060 Jul 16 '17 at 14:34
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    $\begingroup$ You are right but arg($w^2$) could also have been $\frac{\pi}{4} + 2 \pi$ (or more generally $\frac{\pi}{4} + 2 k \pi$ for any integer $k$). This would then give arg($w^2$)$=\frac{9 \pi}{4}$ and your other solution arg($w$)$=\frac{9 \pi}{8}.$ $\endgroup$ – Ronald Blaak Jul 16 '17 at 14:40
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    $\begingroup$ Anyway, I think the problem you have is the line $2\arg(w)=\frac{\pi}{4}\implies \arg(w)=\frac{\pi}{8}$ which isn't true. Actually, $2\arg(w)=\frac{\pi}{4}\implies \arg(w)=\frac{\pi}{8}\lor\arg(w)=\frac{9\pi}{8}$ $\endgroup$ – Shuri2060 Jul 16 '17 at 14:40
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    $\begingroup$ In spite of what you say, you are using “$\sqrt{\,}$” and “arg” as if they are (single-valued) functions on all of $\Bbb C\setminus\{0\}$. Most especially, the rule $\arg(z^n)=n\arg(z)$ does not hold because of the ill-definedness of the argument. $\endgroup$ – Lubin Jul 16 '17 at 14:41
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    $\begingroup$ If it matters, the other value in $(-\pi, \pi]$ is $-\frac{7\pi}{8}$.... $\endgroup$ – Andrew D. Hwang Jul 16 '17 at 14:48

$$\arg(w^2)=\arg(1+i) \implies \arg(w^2)=\frac{\pi}{4}+2k\pi$$ $$\arg(w^n)= n\arg(w)$$ $$\implies \arg(w)=\frac{\pi}{8}+k\pi$$

  • $\begingroup$ So... $\pi<\arg(w)\le \pi\iff k=-1\vee k=0$. Thanks! :) $\endgroup$ – Ixion Jul 16 '17 at 14:48

PRIMER: In general, $\displaystyle \arg(z^n)\ne n\arg(z)$

While it is true that

$$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)\tag 1$$

the equality in $(1)$ is interpreted as a set equivalence. It means that any value of $\arg(z_1z_2)$ can be expressed as the sum of some value of $\arg(z_1)$ and some value of $\arg(z_2)$. And conversely, it means that the sum of any value of $\arg(z_1)$ and any value of $\arg(z_2)$ can be expressed as some value of $\arg(z_1z_2)$.

However, it is not true in general, that $\arg(z^2)=2\arg(2)$.

For example, take $z=i$. Then, for $\arg(z^2)=\arg(-1)=\pi+2n\pi=3\pi$, for $n=1$, there is no value of $2\arg(i)=\pi+4n\pi$ that is equal $3\pi$.

Therefore, $\arg(z^n)\ne n\arg(z)$ in general.

Now, for $z=1+i=\sqrt{2}e^{i(\pi/4+2n\pi)}$, we see that $\arg(z)=\pi/4+2n\pi$ is mulita-valued. And $\arg(\sqrt{z})=\pi/8+n\pi$ is also multi-valued.

If one restricts the argument of $z$ such that $-\pi<\arg(z)\le \pi$, then $\arg(z)$ assumes the values $\pi/4$ and $-3\pi/4$. Finally, the argument of the square root assumes, therefore, the two values $\pi/8$ and $-3\pi/8$.

  • $\begingroup$ Mark viola, maybe you intend $\arg(z^n)\ne n\arg(z)$ $\endgroup$ – Ixion Jul 16 '17 at 14:59
  • $\begingroup$ @Ixion Yes, thank you. I've edited. $\endgroup$ – Mark Viola Jul 16 '17 at 15:01
  • $\begingroup$ Would the cowardly downvoter care to comment? $\endgroup$ – Mark Viola Jul 17 '17 at 13:41

Well, we have when $\text{z}\in\mathbb{C}$:


Where we work in radians, $\left|\text{z}\right|=\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}$, $0\le\arg\left(\text{z}\right)<2\pi$ and $\text{k}\in\mathbb{Z}$

So, when we take the square root of $\text{z}$, we get:

$$\sqrt{\text{z}}=\sqrt{\left|\text{z}\right|\cdot\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)}=\left(\left|\text{z}\right|\cdot\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)\right)^\frac{1}{2}=$$ $$\sqrt{\left|\text{z}\right|}\cdot\left(\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)\right)^\frac{1}{2}=\sqrt{\left|\text{z}\right|}\cdot\exp\left(\frac{\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i}{2}\right)\tag2$$

So, when $\text{z}=w=1+i$ we get:

$$\sqrt{1+i}=\sqrt{\left|1+i\right|}\cdot\exp\left(\frac{\left(\arg\left(1+i\right)+2\pi\text{k}\right)i}{2}\right)=$$ $$\sqrt{\sqrt{2}}\cdot\exp\left(\frac{\left(\frac{\pi}{4}+2\pi\text{k}\right)i}{2}\right)=\sqrt[4]{2}\cdot\exp\left(\left(\frac{\pi}{8}+\pi\text{k}\right)i\right)\tag3$$

And we know that:

  • $$\cos\left(\frac{\pi}{8}+\pi\text{k}\right)=\pm\space\frac{\sqrt{2+\sqrt{2}}}{2}\tag4$$
  • $$\sin\left(\frac{\pi}{8}+\pi\text{k}\right)=\pm\space\frac{\sqrt{2-\sqrt{2}}}{2}\tag5$$
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    $\begingroup$ +1 Thanks, but this is not the answer I'm looking for. :) $\endgroup$ – Ixion Jul 16 '17 at 14:46
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    $\begingroup$ @Ixion I know, but the answer to your question can be found in my answer! $\endgroup$ – Jan Jul 16 '17 at 14:48
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    $\begingroup$ I'm sorry, I did not want to be rough. I know how to calculate the square root of a complex number, but I wanted to avoid making calculations. $\endgroup$ – Ixion Jul 16 '17 at 14:52

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