Number of roots of an equation over complex numbers Given that $\displaystyle x\in \mathbb{C}$ , how many roots does the equation $\displaystyle x=\sqrt{-4} +\sqrt{-1}$ have?
Here are my solving steps
$\displaystyle x=2i+i$
$\displaystyle x=3i$
$\displaystyle x^{2} =-9$
$\displaystyle x^{2} +9=0$
$\displaystyle ( x-3i)( x+3i) =0$
$\displaystyle x=3i\ $and $\displaystyle x=-3i$
I got 2 roots but was marked wrong on an exam and the correct answer is 4 roots. Can someone explain how there are 4 roots? Thank you in advance.
 A: It's not a very well defined question.  It's not clear what $\sqrt{something}$ means.
If $something \ne 0$ there are always two numbers, $x_1$ and $x_2$ so that $x_1^2 = x_2^2 = something$ and, is it turns out, it is always the case that $x_2 = -x_1$.
Now by convention $\sqrt{something}$ is one specific one of these two values.  ONE of them, and not the other one.  The other one is $-\sqrt{something}$
For example $\sqrt{25} = 5$ and $\sqrt {25} \ne -5$ even though $(-5)^2 =5^2 =25$. But it is only $5$ that is equal to $\sqrt{25}$.  $-5  =-\sqrt{25}$.
....
If we use this convention then $\sqrt{-4}=$ is one specific value.  And $\sqrt{-1}$ is one specific value.  And $\sqrt{-4} + \sqrt{-1}$ is one sum and therefore there is one solution to $x = \sqrt{-4} + \sqrt{-1}$.
....
SOmetimes people say $\sqrt{-1} = i$.  and this sets people's teeth on edge.  We define $i$ so that $i^2 = -1$ but we also have $(-i)^2 = i^2 = -i$.  So which value is $\sqrt{-1}$ equal to?  $i$ or $-i$.  Presumably it can't be both.
Well, if people say $\sqrt{-1} = i$ then.... $\sqrt{-1} = i$.  And if we want to be consistant we say if $n > 0$ then then $\sqrt{-n} = i\cdot\sqrt n$.
And if we do this convention:
Then $\sqrt{-4} = i\sqrt{4}=2i$ and $\sqrt{-1} = i$.  An so $x = \sqrt{-4} + \sqrt{-1} = 2i + i = 3i$ is one answer and there is only one answer.
.....
But I have to tell you, no serious mathematician uses the convention.  We know for any $something \in \mathbb C;something\ne 0$ we know that $x^2 = something$ will have two solutions.  One is $x_1$ and the other is $x_2 =-x_1$.  But we have no reason or method to declare one of them over the other as THE square root of $something$.
And so we define $\sqrt{something}$ ONLY when $something \in \mathbb R; something > 0$.  For all other cases we just don't consider $\sqrt {-4}$ to be well defined.
.......
HOWEVER.... if we aske the question.  If $x$ is the sum of A square root of $-4$ plus A square root of $-1$ we figure:  $-4$ has two possible square roots.  One is $2i$ (because $(2i)^2 = 2^2i^2 = 4(-1) = -4$) and the other is $-2i$. But we don't refer to either of them as $\sqrt{-4}$.
And $-1$ has two possible square roots: $i$ and $-i$.
So the possible values are
$x=\begin{cases}2i + i = 3i\\-2i + i = -i\\2i -i = i\\-2i -i = -3i\end{cases}$
There are four possible values for $x$.
IF that was the question that was being asked. But the isn't the way the question was written.  The way the question was written is unclear.
A: There are two numbers that square to $-4$: $\pm 2\mathrm{i}$.  There are two numbers that square to $-1$: $\pm \mathrm{i}$.  Consequently, all of \begin{align*}
x &= \pm 2 \mathrm{i} \pm \mathrm{i}  \\
    &= \{2 \mathrm{i} + \mathrm{i}, 2 \mathrm{i} - \mathrm{i}, -2 \mathrm{i} + \mathrm{i}, -2 \mathrm{i} - \mathrm{i}\}  \\
    &= \{ 3\mathrm{i}, \mathrm{i}, -\mathrm{i}, -3\mathrm{i} \}
\end{align*}
are roots.
A: Squaring both sides introduces extraneous solutions.  An $n$-th degree polynomial over $\Bbb C$ has $n$ roots, by the fundamental theorem of algebra.
Therefore there is one solution.
