Complex coefficient requiring the polar form. z2 = 3 + 4i

The complex number is $z^2 = 3+4i$.
First, find the modulus $r$:

$$r = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$$

Next, find the argument $\theta$. Since the real part (3) and imaginary part (4) are both positive, $\theta$ is in the first quadrant.

$$\theta = \arctan\left(\frac{4}{3}\right)$$

So, $z^2$ in polar form is $5\left(\cos\left(\arctan\left(\frac{4}{3}\right)\right) + i \sin\left(\arctan\left(\frac{4}{3}\right)\right)\right)$.
Let $\alpha = \arctan\left(\frac{4}{3}\right)$. Then $z^2 = 5(\cos \alpha + i \sin \alpha)$.

Let $z = \rho(\cos \phi + i \sin \phi)$. Then $z^2 = \rho^2(\cos(2\phi) + i \sin(2\phi))$.
Comparing this with $z^2 = 5(\cos \alpha + i \sin \alpha)$:

$$\rho^2 = 5 \implies \rho = \sqrt{5}$$

And for the argument:

$$2\phi = \alpha + 2k\pi$$ $$\phi = \frac{\alpha}{2} + k\pi$$

where $k$ is an integer. We need to find two distinct roots, so we use $k=0$ and $k=1$.

From $\alpha = \arctan\left(\frac{4}{3}\right)$, we know that $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$ (from a 3-4-5 right triangle).
Using the half-angle identities:

$$\cos^2\left(\frac{\alpha}{2}\right) = \frac{1 + \cos \alpha}{2} = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{4}{5}$$ $$\sin^2\left(\frac{\alpha}{2}\right) = \frac{1 - \cos \alpha}{2} = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5}$$

Since $\alpha$ is in the first quadrant ($0 < \alpha < \frac{\pi}{2}$), $\frac{\alpha}{2}$ is also in the first quadrant ($0 < \frac{\alpha}{2} < \frac{\pi}{4}$). Therefore, both $\cos\left(\frac{\alpha}{2}\right)$ and $\sin\left(\frac{\alpha}{2}\right)$ are positive.

$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$ $$\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$$

For $k=0$:

$$\phi_0 = \frac{\alpha}{2}$$ $$z_0 = \rho(\cos \phi_0 + i \sin \phi_0) = \sqrt{5}\left(\frac{2}{\sqrt{5}} + i \frac{1}{\sqrt{5}}\right)$$ $$z_0 = 2 + i$$

For $k=1$:

$$\phi_1 = \frac{\alpha}{2} + \pi$$ $$z_1 = \rho(\cos \phi_1 + i \sin \phi_1) = \sqrt{5}\left(\cos\left(\frac{\alpha}{2} + \pi\right) + i \sin\left(\frac{\alpha}{2} + \pi\right)\right)$$

Using the identities $\cos(\theta+\pi) = -\cos\theta$ and $\sin(\theta+\pi) = -\sin\theta$:

$$z_1 = \sqrt{5}\left(-\cos\left(\frac{\alpha}{2}\right) - i \sin\left(\frac{\alpha}{2}\right)\right)$$ $$z_1 = \sqrt{5}\left(-\frac{2}{\sqrt{5}} - i \frac{1}{\sqrt{5}}\right)$$ $$z_1 = -2 - i$$

The two square roots of $3+4i$ are $\boxed{2+i \text{ and } -2-i}$.