Can you apply de-moivre's

$\begin{aligned} \left(i-\sqrt{3}\right)^{13} & = X\left(i-\sqrt{3}\right) \\ \Rightarrow X & = \left(i-\sqrt{3}\right)^{12} \\ & = \left[2 \left( -\frac{\sqrt{3}}{2} + \frac{1}{2}i \right)\right]^{12} \\ & = 2^{12} \left( e^{\frac{5\pi}{6}i} \right)^{12} \\ & = 2^{12} e^{10\pi i} \\ & = 2^{12} = \boxed{4096} \end{aligned}$

The more simple solution is to apply the absolute value. Given that $(i-\sqrt{3})^{13} = x(i-\sqrt{3})$ Applying the absolute value to both sides will give $|i-\sqrt{3}|^{13} = |x(i-\sqrt{3})|$ , or $2^{13} = 2|x|$ , or simply $|x| = 2^{12} = 4096$ . Given that $arg(i-\sqrt{3}) = \frac{5\pi}{3}$ , $arg((i-\sqrt{3})^{12}) = 20\pi\ > 0$ , therefore $x$ shall be positive, and $x = 4096$

$(i-\sqrt{3})^{13}=(i-\sqrt{3})^{12}(i-\sqrt{3})=2^{12}cis^{12}(\frac{\pi}{3})(i-\sqrt{3})$ $=4096cis(4\pi)(i-\sqrt{3})=4096(i-\sqrt{3})$

Notice that:

$i-\sqrt{3}=2(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)=2(\cos 150°+i\sin 150°)$

which is twice the 12th root of unity. Then:

$X=(i-\sqrt{3})^{12}=4096$