Find the complex square root of $$7+6\sqrt2i$$ giving your answer in the form $x+iy$ where $x$ and $y$ are real.
The answer I have gotten is $-23+84\sqrt2i$ by squaring $7+6\sqrt2i$, but the “correct” answer is $\pm(3+i\sqrt2)$.
Can you explain how I got this wrong?
The fast way to obtain square roots of complex numbers is this: by identification of $(x+iy)^2$ with the given number, you obtain the non-linear system \begin{cases} x^2-y^2=7,\\ xy=3\sqrt 2. \end{cases} You may add the relation deduced from comparing $|x+iy|^2$ with the given number modulus: $$x^2+y^2=\sqrt{49+72}=11,$$ so that you obtain a very simple linear system in $x^2$ and $y^2$: \begin{cases} x^2-y^2=7,\\ x^2+y^2=11. \end{cases} You can solve this system and deduce the values of $x$ and $y$, using the third equation, which implies that $x$ and $y$ have the same sign.
A different method for rendering complex square roots with non-real radicands is what I call "the angle bisection method", from a familiar geometric construction.
Render
$(z+|z|)^2=z^2+2z|z|+|z|^2=z^2+2z|z|+z\overline{z}$
$=2z|z|+2z(z+\overline{z})$
$\color{blue}{=2z(|z|+\Re(z))}$
So, upon taking square roots and solving:
$\sqrt z=\pm(z+|z|)/\sqrt{2(|z|+\Re(z))}$
Note the denominator is real, so the real and imaginary parts are automatically resolved. Here $|z|=11$ is rational, allowing a simplified form for the square root. We directly get, as given in other answers, $\pm (3+\sqrt 2i)$.
Let square roots of $7+6\sqrt{2}i $ be $\pm(a+bi);$ squaring this we have $a^2-b^2+2abi=7+6\sqrt{2}i;$ comparing coefficients we have $a^2-b^2=7,ab=3\sqrt{2};$ substituting for $b$ in the first equation we have $a^2-\frac{18}{a^2}=7;$ multiplying throughout by $a^2$ and rearranging we get $a^4-18-7a^2=a^4+2a^2-9a^2-18=a^2(a^2+2)-9(a^2+2)=(a^2-9)(a^2+2)=0$ thus $a=\pm 3$ (as $a$ is a real number in our assumption we rule out the second case, i.e., $a=\pm \sqrt{2}i)$ now finding out $b=\pm \sqrt{2}$. Thus the square roots are $\pm(3+\sqrt{2}i)$
$$\sqrt{7+6\sqrt2i}=\sqrt{9+6\sqrt2i+(\sqrt2i)^2}=\sqrt{(3+\sqrt2i)^2}=\{3+\sqrt2i,-3-\sqrt2i\}.$$ By the way, $$(-23+84\sqrt2i)^2\neq7+6\sqrt2i.$$
Maybe you calculated $(7+6\sqrt2i)^2$?
Let $z = x + yi$ is a complex number where $x,y \in \mathbb{R}$, and let $\theta = \arg(z) \in [0,2\pi] \subseteq \mathbb{R}$. The square root
$$\begin{aligned} \sqrt{z} &= \sqrt{r\cos \theta + ir\sin \theta} = \sqrt{r}\cos \frac{\theta}{2} + i\sqrt{r}\sin \frac{\theta}{2} \\ &= \sqrt{r}\sqrt{\frac{1 + \cos \theta}{2}} + i\sqrt{r}\sqrt{\frac{1 - \cos \theta}{2}} \\ &= \sqrt{\frac{|z| + \Re(z)}{2}} + i\sqrt{\frac{|z| - \Re(z)}{2}}. \end{aligned}$$
Hence $$\sqrt{7 + i6\sqrt{2}} = \sqrt{\frac{11 + 7}{2}} + i\sqrt{\frac{11 - 7}{2}} = 3 + i\sqrt{2}.$$
An easy way to do this is to convert to polar form and then use De Moivre, but the way the question is framed seems to want you to do it the straightforward way, namely square $x+iy$ and equate it to $7+6\sqrt 2 i,$ which yields the $2×2$ quadratic system $$x^2-y^2=7, 2xy=6\sqrt 2.$$ Since both $\text{LHS}$ are homogeneous of order two, a substitution will yield a biquadratic either in $x$ or $y,$ which is easily solved.
{2}instead of regular brackets, $\sqrt{2}$ renders correctly. In fact, since $2$ is a single digit, you don't really need brackets at all:\sqrt 2works just fine. – Arthur Jun 21 '19 at 08:49