# Evaluating $\sqrt{a\pm bi\sqrt c}$

I recently encountered this problem $$\sqrt{10-4i\sqrt{6}}$$ To witch I set the solution equal to $$a+bi$$ squaring both sides leaves $${10-4i\sqrt{6}}=a^2-b^2+2abi$$ Obviously $$a^2-b^2=10$$ and $$2abi=-4i\sqrt{6}$$, using geuss and check, the solution is $$a=\sqrt12, b=\sqrt2$$

But I was wondering if there is a faster way to solve these types of problems or a method that doesn't involve guess and check (since it can get tedious at times) for the basic form $$\sqrt{a\pm bi\sqrt c}=x+yi$$ Where you're solving for $$x$$ and $$y$$.
I've attempted but failed since it gets pretty messy. All formula will be very much appreciated. (not all equations of that form can reduce)

• "solution" usually means finding an unknown. What is the unknown here? Jul 25, 2019 at 22:28
• your right i should have said formula
– user610551
Jul 25, 2019 at 22:29
• $a^2-b^2=10$ and $2ab=-4\sqrt6\implies a^2=\dfrac{24}{b^2}$ and $\dfrac{24}{b^2}-b^2=10\implies b^4-10b^2+24=0\implies(b^2-2)(b^2-12)=0$ Jul 25, 2019 at 22:30
• Now do you think its possible to solve the form $\sqrt{a\pm bi\sqrt c}$ where $a,b,c$ are intergers
– user610551
Jul 25, 2019 at 22:32
• One thing right off the bat. Every non-zero complex number will have two square roots so there wont be just $x + yi$ there will also be $-x -yi$ Jul 25, 2019 at 23:23

One thing right off the bat. Every non-zero complex number will have two square roots so there wont be just $$x + yi$$ there will also be $$-x -yi$$

" using geuss and check" Why use guess and check when you can use the quadratic formula?

$$a^2 - b^2 = 10$$ and $$2abi = -4\sqrt 6i$$

$$a=\frac{-2\sqrt 6}b$$

$$\frac {4*6}{b^2} - b^2 = 10$$

$$24 - b^4 = 10b^2$$

$$b^4 + 10b^2 - 24 = 0$$

$$b^2 = \frac {-10 \pm \sqrt{100+96}}2= -5 \pm \sqrt{25+ 24}=-5\pm 7$$.

As $$b$$ is presumed to be real $$b^2 = 2$$ and $$b=\pm \sqrt 2$$

And $$a=\frac{-2\sqrt 6}b= \mp \frac {2\sqrt 6}{\sqrt{2}}=\pm 2\sqrt 3(= \pm \sqrt {12})$$

So $$\sqrt{10-4\sqrt 6i} = \pm(-2\sqrt 3 + \sqrt 2i)$$

No guessing. No checking.

For general $$z\in\mathbb{C}$$, the expression $$\sqrt{z}$$ means the principal square root of $$z$$.

Assuming that $$a,b,c\in\mathbb{R}$$ and $$c\ge 0$$, one can write $$w:=a+ bi\sqrt{c}=re^{i\theta}$$ for some $$r\ge 0$$ and $$\theta\in(-\pi,\pi]$$ such that $$\sqrt{a^2+b^2c}=r,\quad r\cos\theta = a,\quad r\sin\theta =b\sqrt{c}.$$

It then follows that $$\sqrt{w}=\sqrt{r}e^{i\theta/2}.$$

Now you use Euler's formula. Finding $$r$$ is straightforward. One needs the inverse trigonometric functions for $$\theta$$ in general.

[Added:] If you only consider the general case for $$a,b,c\in\mathbb{R}$$ and $$c\ge 0$$, there is nothing special about $$\sqrt{c}$$ and one can simply consider $$w=a+ib$$ by introducing a new variable. In this case $$r = \sqrt{a^2+b^2},\quad \tan\theta = \frac{b}{a},\quad \theta\in(-\pi,\pi].$$

[Added later:] In the Wikipedia article mentioned above, one can see the algebraic formula:

When the (complex) number is expressed using Cartesian coordinates the following formula can be used for the principal square root: $${\sqrt {x+iy}}={\sqrt {\frac {{\sqrt {x^{2}+y^{2}}}+x}{2}}}\pm i{\sqrt {\frac {{\sqrt {x^{2}+y^{2}}}-x}{2}}},}$$ where the sign of the imaginary part of the root is taken to be the same as the sign of the imaginary part of the original number, or positive when zero. The real part of the principal value is always nonnegative.

We have that $$a^2-b^2=10 \quad \textrm{and} \quad 2ab=-4\sqrt 6$$ Now, squaring both equalities and addem up we get $$(a^2+b^2)^2=(a^2-b^2)^2 +(2ab)^2=10^2+(-4\sqrt 6)^2 =196$$ $$\Rightarrow \quad a^2+b^2=14$$ and using again the first equality we obtain $$a^2=12 \quad \textrm{and} \quad b^2=2$$ or $$a=\pm 2\sqrt 3 \quad \textrm{and} \quad b=\pm \sqrt 2$$ but $$ab<0$$ (by the second equality), that is, $$a$$ and $$b$$ are the opposite sign. Thus, the solutions are given by $$a= 2\sqrt 3 \quad \textrm{and} \quad b=- \sqrt 2$$ or well $$a=-2\sqrt 3 \quad \textrm{and} \quad b=\sqrt 2$$

• thanks, I was asking if you could give a formula for the solution in the form $\sqrt{a\pm bi\sqrt c}$ (like a quadratic formula)
– user610551
Jul 25, 2019 at 22:35
• Yes, but more general (the $\pm$ and the $\sqrt c$ could be omitted) only considering $\sqrt{a+bi}$ : $$\pm \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} \pm i \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}$$ (the signs depend but can be determined as in the analysis I did in my answer). Jul 25, 2019 at 22:50
• thank you, would you be kind enough to show a step by step answer
– user610551
Jul 25, 2019 at 22:52
• And the proof of this formula is essentially the same as what I already did. I'm glad to have helped you. Jul 25, 2019 at 22:58

Concrete your problem we can solve so: $$\sqrt{10-4i\sqrt6}=\sqrt{12-4i\sqrt6-2}=\sqrt{(2\sqrt3-\sqrt2i)^2}=\pm(2\sqrt3-\sqrt2i).$$

• although that is very elegant, it'll take a long time to come up with when in competition. But thanks!!
– user610551
Jul 26, 2019 at 2:52
• @Leonard Euler I always try before to find a similar solutions. Jul 26, 2019 at 2:54