Determine the number of zeros in the first quadrant $f(z) = z^4- 3z^2 + 3$ Determine the number of zeroes of the following function which are in the first quadrant:
$$f(z) = z^4- 3z^2 + 3$$
Help please!!! I'm not that good at complex variables!
 A: You can see that if $z_0$ is a zero then so is $-z_0$, $\overline{z_0}$, $-\overline{z_0}$.
It is straightforward to show that the restriction of $f$ to the real axis has a minimum $f({3 \over 2}) = {3 \over 4}$, so $f$ has no real axis zeros.
If $f$ had an imaginary axis zero, then $f$ would have the form $f(z) = z^4 + c^2$ for some real $c$, hence $f$ has no imaginary axis zeros.
Since $f$ has four zeros of the form $z_0$, $-z_0$, $\overline{z_0}$, $-\overline{z_0}$, it follows that each quadrant has exactly one zero.
A: Hint: Substitute $w = z^2$ and use quadratic formula.
Hint 2: If a root is in the first quadrant, then both its real and imaginary parts are positive.
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Following the hint, we have $w^2 - 3w + 3 = 0$. so $w = \frac{3 \pm i\sqrt{3}}{2}$ and $z = \pm \sqrt{\frac{3 \pm i\sqrt{3}}{2}}$
The root $\frac{3 + i \sqrt3}{2}$ is in the first quadrant with some angle $\theta$ between $0$ and $90$, thus when we take the square root this angle will be halved -- it will still be in the first quadrant. However, the negative iteration will be in the third quadrant.
The root $\frac{3 - i \sqrt{3}}{2}$ is in the fourth quadrant (as well as its positive square root), while its negative square root is in the second quadrant.
Thus there is only one root in first quadrant.
A: set $w=z^2$ then solve the quadratic equation $w^2-3w+3$. You will get:
$$w = \dfrac{3 \pm \sqrt{9-4(1)(3)}}{2} = \dfrac{3\pm\sqrt{3}i}{2}$$ 
So now:
$$z^2 = \dfrac{3\pm\sqrt{3}i}{2}$$
A root is in the first quadrant when the real and imaginary parts are positive. When you take the square root of $z^2$ this will leave you with one solution in the first quadrant.
