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Since any complex number can be of polar form. We set that $z = r\cos \theta +ir\sin \theta$. Now by de Moivre's Theorem, we easily see that $$z^{4} = r^{4}\cos4 \theta + ir^{4}\sin4 \theta$$

Since $$z^{4} = -1+\sqrt{3}i$$

We equip accordingly and see that $$r^{4}\cos4 \theta = -1$$ $$r^{4}\sin4 \theta = \sqrt{3}$$

Solving the above 2 equations we have $$\tan 4\theta = -\sqrt{3} \Rightarrow 4\theta = -\dfrac{\pi}{3} \Rightarrow \theta = -\dfrac{\pi}{12}$$

Hence we have $$\text{arg}z = -\dfrac{\pi}{12} + k\pi$$

Furthermore, $$\text{Arg}z = -\dfrac{\pi}{12} \text{ or } \dfrac{11\pi}{12}$$

However, i cannot find out what $r$ is, as when i substitute in to solve, $r$ become a complex number?? Is my answer correct, how can i make my steps better? Thanks!

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Observe that $-1+\sqrt{3}i=2e^{\frac{2\pi i}{3}}$. Then the roots of the equation are $$ z=\sqrt[4]{2}e^{\frac{\pi i}{6}+\frac{k\pi}{2}}, $$ where $k=0,1,2,3$. Thus $r=\sqrt[4]{2}$. Substitute $k$, then we get $z=\sqrt[4]{2}e^{\frac{\pi i}{6}}$, $z=\sqrt[4]{2}e^{\frac{2\pi i}{3}}$, $z=\sqrt[4]{2}e^{\frac{7\pi i}{6}}$, and $z=\sqrt[4]{2}e^{\frac{5\pi i}{3}}$. There is no root whose argument is $\frac{11\pi}{12}$. What happened? Because your attempt has a flaw: Since $r^4 \cos 4\theta=-1$ and $r^4\sin 4\theta=\sqrt{3}$, we get $\tan 4\theta=-\sqrt{3}$, but it doesn't imply that $4\theta = -\frac{\pi}{3}$. If it is true, then $r^4 \cos4\theta$ cannot be negative!

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    Am i right to rectify it as follows : Since $-1 + \sqrt{3} i$ is in the 2nd quadrant, we must have $4\theta = 2\pi/3$? @choco_addicted – 2017-01-15
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    @ilovewt I believe so. – 2017-01-15
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    @ilovewt. It is always a good practice to graph the complex number to see in which quadrant it lies. Then you can use the arctan to get the argument and add 180 if needed for quadrant 2 or 3...That way you can't go wrong. Question: What argument would you assign if the complex number happens to be on the imaginary axis? – 2017-01-15
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    @imranfat should it be 90 degrees? – 2017-01-15
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    @ilovewt. No, it is 180. The arctan on your calculator produces values between -180 and 180 and spans the 4th and 1st quadrant. So in order to get an angle in the 2nd and 3rd quadrant, you'll have to add 180 to get into the right quadrant! – 2017-01-16
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    @ilovewt. I meant to say, the arctan on your calculator produces values between -90 and 90. So for quadrant 1 and 4, it is ok. If the complex number is in quadrant 2, your calculator gives you the argument in quadrant 4 so you need to add 180. If the complex number is in quadrant 3, the calculator gives you an angle in quadrant 1. Hence the 180 addition... – 2017-01-17
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    @imranfat You mentioned that the complex number happens to be on the imaginary axis, which by convention should be the y-axis. Hence the complex number is of the form $z = 0+iy$ and if $y >0$, then the angle should be 90 degrees, if $y<0$ , then the angle should be -90 degrees? – 2017-01-21
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    @ilovewt. Yes!! – 2017-01-22
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The absolute value is very simple:

$$|z^4|=|-1+\sqrt3 i|=2$$

Thus,

$$|z|=\sqrt[4]2$$