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I need to compute the square root of $3 + 2\sqrt{10}i$.

I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:

$$ \sqrt{3 + 2\sqrt{10}i} = x + iy \quad \longrightarrow \quad 3 + 2\sqrt{10}i = x^2 - y^2 +2xyi $$ and so forth, but my answer isn't correct.

Frenzy Li
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jimpix
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  • BTW, there are two square roots. So using 'the square root' isn't quite clear. – Zain Patel Aug 06 '16 at 12:24
  • By Pythagoras $|3+2\sqrt{10}i|=\sqrt{49}=7$. So the square root has absolute value $\sqrt7$. In this case the numbers are simple, so if we expect a "nice" result (a big if!!) the obvious thing to check is $$(\sqrt5+\sqrt2 i)^2=\cdots$$ A big tip-off is that $3=(\sqrt5)^2-(\sqrt2)^2$. – Jyrki Lahtonen Aug 06 '16 at 13:45
  • well, what is your answer and how did you get it? We can't tell you what you did wrong if we don't know what you did. And we are certainly not going to just give you the right answer. So far you did every thing right. $x^2 - y^2 = 3$ and $xy = \sqrt{10}$. So how can we possibly help you further if we don't know what you did next? – fleablood Aug 06 '16 at 16:16

6 Answers6

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As I explained here. there is a very simple formula for denesting such radicals, namely

Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

$\ 3+2\sqrt{-10}\ $ has norm $= 49.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 7\,\ $ yields $\ {-}4+2\sqrt{-10}\:$

with $\, {\rm\ \sqrt{trace}}\, =\, 2\sqrt{-2}.\ \ \ \rm \color{brown}{Dividing\ this\ out}\ $ of the above we obtain $\,\ \sqrt{-2} + \sqrt 5$

Checking it we find $\,\ (\sqrt{-2} + \sqrt 5)^2 =\, -2+5 + 2\sqrt{-2}\sqrt 5\, =\, 3+ 2\sqrt{-10}$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.

Community
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Bill Dubuque
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    Nothing wrong with your main point. Yet you should keep in mind that beginners are not always aware of the perils of using multivalued square roots such as $\sqrt{-2}$ and $\sqrt{-10}$. Also carelessly assuming that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ leads to those famous fake proofs of $-1=1$ et cetera. – Jyrki Lahtonen Aug 06 '16 at 13:49
  • @JyrkiLahtonen I imagine one just has to be consistent about a choice of $\sqrt{-1}$? – snulty Aug 06 '16 at 15:54
  • Correct @snulty. For the record: I know that Bill won't fumble a simple thing like this. I was just a bit surprised that he would leave the branch choice hanging in the air. Particularly as the OP used the form $\sqrt{10}i$. Anyway, the denesting technique is nice, and if Bill wishes to write it this way that's his decision. – Jyrki Lahtonen Aug 06 '16 at 16:01
  • @JyrkiLahtonen It is not clear if the question is algebraic or analytic. The method is algebraic, so of course you need to pay attention to (principal) branches if you want to interpret the result analytically. – Bill Dubuque Aug 06 '16 at 16:10
  • So you seek to give a solution without taking a position on the choice of branch. That is economical. – Jyrki Lahtonen Aug 06 '16 at 16:16
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Your "and so on" could go like this:

$$\begin{cases}x^2-y^2=3,\\2xy=2\sqrt{10}.\end{cases}$$

Then squaring and adding both,

$$x^4-2x^2y^2+y^4+4x^2y^2=(x^2+y^2)^2=49,$$ so that

$$x^2+y^2=\pm7.$$

Solving with the help of the first,

$$x^2=5,y^2=2\text{ or }x^2=-2,y=-5.$$

This leaves the possibilities

$$x=\pm\sqrt5,y=\pm\sqrt2.$$

By the second equation, we know the signs are synchronized, hence

$$\sqrt5+i\sqrt2\text{ or }-\sqrt5-i\sqrt2.$$

More generally,

$$\begin{cases}x^2-y^2=u,\\2xy=v.\end{cases}$$

yields

$$u^2=\frac12(\sqrt{v^2+u^2}+u),v^2=\frac12(\sqrt{v^2+u^2}-u),$$

and

$$x=\pm\sqrt{\frac12(\sqrt{v^2+u^2}+u)}, y=\pm\sqrt{\frac12(\sqrt{v^2+u^2}-u)},$$

where the sign of $xy$ must match the sign of $v$.

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$z=3+2i\sqrt{10}$ is a complex number with modulus $7=\sqrt{9+40}$ in the first quadrant ($\text{Re}(x),\text{Re}(y)>0$), hence we have $$ z=7 e^{i\theta} $$ with $$\theta = \arctan\frac{2\sqrt{10}}{3}=2\arctan t.$$ Since $\tan(2u)=\frac{2\tan(u)}{1-\tan^2(u)}$, by solving $\frac{2t}{1-t^2}=\frac{2\sqrt{10}}{3}$ with the constraint $t>0$ we get $t=\sqrt{\frac{2}{5}}$, hence the square root of $z$ in the first quadrant is given by: $$ \sqrt{z} = \sqrt{7} e^{it} = \sqrt{7}\left(\cos\arctan\sqrt{\frac{2}{5}}+i\sin\arctan\sqrt{\frac{2}{5}}\right) $$ or: $$ \sqrt{z} = \sqrt{7}\left(\sqrt{\frac{5}{7}}+i\sqrt{\frac{2}{7}}\right) = \color{red}{\sqrt{5}+i\sqrt{2}}. $$ Let we check the solution so found: $$ (\sqrt{5}+i\sqrt{2})^2 = 5-2+2i\sqrt{2\cdot 5} = 3+2i\sqrt{10}.$$

Jack D'Aurizio
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    I wanted to post an answer like this with cos and sin of arctan, but I got tripped but by the $2\arctan$! That's very instructive, thank you for a good answer! (+1) – snulty Aug 06 '16 at 15:49
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    It might be slightly clearer to say why you introduce $t$ before doing so: you're writing the given $z$ as $7e^{i \theta}$ and then writing $\sqrt{z}$ as $\sqrt{7}e^{i \theta/2}$, denoting $\theta/2$ by $\arctan(t)$. It took me a moment to figure that out. – Ian Aug 06 '16 at 16:13
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Just using calculus, you are looking for the number $z=x+iy$ such that $z^2=3+2\sqrt{10}i$. Expand $$z^2=x^2-y^2+2ixy=3+2\sqrt{10}i$$ Identify the real and imaginary parts; this gives $$x^2-y^2=3\tag 1$$ $$xy=\sqrt{10}\implies y=\frac {\sqrt{10}}x\tag 2$$ So, equation $(1)$ becomes $$x^2-\frac {10}{x^2}=3\implies x^4-3x^2-10=0\tag 3$$ Which is a quadratic equation in $x^2$; let $t=x^2$ and solve $t^2-3t-10=0$ the roots of which are $??$. One solution must be discarded since the product of the roots is negative; so, only one $t$ must be kept. Now, back to $x^2=t$ will give $x$ and $(2)$ will give the corresponding $y$.

Claude Leibovici
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$$ \sqrt{3 + 2\sqrt{10}i} = x + iy $$ $$ 3 + 2\sqrt{10}i=x^2 -y^2 +2ixy $$ $$ x^2 -y^2 =3$$ $$ 2xy=2 \sqrt{10}$$ $$( x^2 + y^2 )^2= ( x^2 - y^2 )^2 +2xy $$ $$ x^2 + y^2 =7$$ $$ x^2 -y^2 =3$$

Aakash Kumar
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You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that

$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$

Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:

$$\pm(\sqrt{5}+\sqrt{2}i).$$

smcc
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