While doing a certain problem based on complex numbers I faced this doubt.
When $\sqrt{2i}$ is mentioned in a question should I take it's value as $(1+i)$ or both $(1+i)$ and $(-1-i)$ ?I mean should I take only the positive root or both the roots?
Well this is a very basic level doubt but I think its good to have clear concepts :-P
Note: the same things said here apply to any multi-valued function $z \in \Bbb C^* \mapsto z^c$, where $c \in \Bbb C$.
We have for $z \in \Bbb C^*$,
$$\sqrt z := \exp(\frac12 \log z)$$
where $\log$ is the multi-valued logarithm function. So, the correspondence $z \mapsto \sqrt z$ is not a function, but a multi-valued function.
The multi-valued logarithm function is defined, on $\Bbb C^*$, as $\log z = \ln|z| + i \arg z$, where $\arg$ is the multi-valued argument function.
By Euler's formula, each non-zero complex number $z$ can be written as:
$$z = r(\cos \theta + i \sin \theta)$$
for some unique $r > 0$, called the modulus of $z$, and for some unique (up to addition of $2\pi$) real number $\theta$. The multi-valued argument function is defined for such a $z$ by $\arg z = \{\theta + 2k \pi, k \in \Bbb Z\}$.
It is a basic result that for any $\alpha \in \Bbb R$, $\arg$ assigns to any non-zero complex number a unique value in $[\alpha, \alpha + 2\pi)$. So, the correspondence $\arg_{\alpha}$, which associates to each $z \in \Bbb C^*$ the unique value of $\arg$ in $[\alpha, \alpha + 2\pi)$, is a function. Note that any such $\arg_{\alpha}$ is continuous on $\Bbb C$ except on the ray $\{z = |z|\exp(i\alpha)\}$. The principal argument is defined as $\text{Arg} = \arg_{-\pi}$ (some define it as $\arg_{0}$ and some even take $(\alpha, \alpha + 2\pi]$; these are just technical issues).
For any $\alpha$, $\arg_{\alpha}$ gives rise to a $\log_{\alpha}$ defined in the obvious way: $\ln|z| + i \arg_{\alpha} (z)$. So, we finally get a true function, call it $P_{\alpha}$, defined by $P_{\alpha}(z) = \exp(\frac12 \log_{\alpha} z)$.
Had we chosen $\alpha = -\pi$, we would have got:
$$\sqrt{2i} = P_{\alpha}(2i) = \exp(\frac12(\ln|2i| + i \text{Arg}(2i)) = 1 + i$$
and had we chosen $\alpha = \pi$, we would have got:
$$\sqrt{2i} = P_{\alpha}(2i) = \exp(\frac12(\ln|2i| + i \arg_{\alpha}(2i)) = -(1 + i)$$
This is why we can't meaningfully just talk about the square root of a complex number, but we have to specify an $\alpha$.
Usually, though, $\sqrt z$ is interpreted as $P_{- \pi}$.