# Modulus of complex numbers

$z_1, z_2$ and $z_3$ are complex numbers related by:

$$z_3=\sqrt{\frac{z_1}{z_2}}$$

When I take the modulus (defined for a complex number)of the L.H.S, how do I reflect that on the R.H.S? Am I allowed to directly take the modulus of the complex numbers on the R.H.S separately and write this:

$$|z_3|=\sqrt{\frac{|z_1|}{|z_2|}}\ ?$$

• Somebody should mention that $$\sqrt{\frac{|z_1|}{|z_2|}}$$ exists but $$\sqrt{\frac{z_1}{z_2}}$$ does not. (Unrelated: 1. Please replace each \cfrac by \frac. 2. There is no "conjugate" in the question, hence why the title?) – Did Aug 18 '16 at 10:36
• Oh, sorry, my bad! – user361896 Aug 18 '16 at 10:49

By definition, $$z_3^2=\frac{z_1}{z_2}$$ (with the assumption that $z_2\ne0$) and therefore $$|z_3|^2=|z_3^2|=\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$$ because $|uv|=|u|\,|v|$ for $u,v\in\mathbb{C}$. Since the final equalities are between nonnegative real numbers, we conclude that $$|z_3|=\sqrt{\frac{|z_1|}{|z_2|}}$$

• But how is |z|²=|z²|? – user361896 Aug 18 '16 at 10:46
• @KaumudiHarikumar $|u|^2=u\bar{u}$ by definition. So $|uv|^2=uv\overline{uv}=(u\bar{u})(v\bar{v})=|u|^2|v|^2$. Taking square roots, we get $|uv|=|u|\,|v|$ and, in particular, $|z^2|=|z|^2$ (with $z=u=v$). – egreg Aug 18 '16 at 10:53
• Oh crap! My bad! I'm so sorry I made u type all that. I figured it out on my own. Thanks! :) – user361896 Aug 18 '16 at 10:55

$$\text{z}_3=\sqrt{\frac{\text{z}_1}{\text{z}_2}}\to\left|\text{z}_3\right|=\left|\sqrt{\frac{\text{z}_1}{\text{z}_2}}\right|=\left|\left(\frac{\text{z}_1}{\text{z}_2}\right)^{\frac{1}{2}}\right|=\left(\left|\frac{\text{z}_1}{\text{z}_2}\right|\right)^{\frac{1}{2}}=\left(\frac{|\text{z}_1|}{|\text{z}_2|}\right)^{\frac{1}{2}}=\sqrt{\frac{|\text{z}_1|}{|\text{z}_2|}}$$

• How did u write the last step?! Is there some property that I'm unaware of? – user361896 Aug 18 '16 at 9:51
• Something to the power of 1/2 means the quare root of that something. This are basic properties of finding a complex number's modulus. You can prove those with the complex exponentials – Jan Aug 18 '16 at 9:55
• Oh, so can you help me to prove it? Am I correct in assuming that this is the case for all exponents? That regardless of the way multiple complex numbers are connected within an expression, we simply take the mod of all of them? – user361896 Aug 18 '16 at 9:59
• You can use those for all complex numbers that are in these form! Of you look to the post of N.H. he proved some of those rules – Jan Aug 18 '16 at 10:09
• Sorry but there is no canonical function $\sqrt{\ }:\mathbb C\to\mathbb C$, so what does $\sqrt{z}$ mean in your answer? Anyway, your first comment in the present thread is squarely misleading. – Did Aug 18 '16 at 10:40

Hint : you can use exponential form of a complex number : $z = re^{i\theta}$.

Edit : here is more details.

Write your complex numbers as $z_1 = r_1e^{i\theta_1}, r_2e^{i\theta_2}$ and $r_3e^{i\theta_3}$. Notice that $\sqrt z_1 = \sqrt r_1 e^{i\theta_1/2}$, $\sqrt{\frac{z_2}{z_3}} = \sqrt{\frac{r_2}{r_3}} e^{i(\theta_2 - \theta_3)/2}$. Since $r_i = |z_i|$ the last equality is what we want.

Edit 2 : notice that as Did mentionned, there is no well defined notion of a square root of a complex number. So the usually solution is to define the square root of a complex numbers $z \in \mathbb C \backslash \mathbb R_-$ by the formula above.

• Umm, can u please help me some more? How should I proceed to arrive at what Jan Eerland has "proved"? Are there some properties that I'm unaware of? – user361896 Aug 18 '16 at 9:50
• Sorry but there is no canonical function $\sqrt{\ }:\mathbb C\to\mathbb C$, so what does $\sqrt{z}$ mean in your answer? – Did Aug 18 '16 at 10:39
• Sheesh! I'm an idiot for not having been able to do this on my own! Thanks so much sir/ma'am! :) – user361896 Aug 18 '16 at 10:40
• Did : since it was not precised in the question I did assume the author took any determination of the square root on some subset of $\mathbb C$. You're right I should be more careful about it. – user171326 Aug 18 '16 at 10:42
• @KaumudiHarikumar : you're welcome :) – user171326 Aug 18 '16 at 10:45