Angle by which tangents to curves at $z_{0}$ are rotated under the mapping $w = z^{2}$

I have to find an angle by which tangents to curves at $z_{0}$ are rotated under the mapping $w = z^{2}$ if (a) $z_{0} = i$, (b) $z_{0} = -1/4$, (c) $z_{0} = 1+i$, and also find the corresponding values of magnification.

I know that if $z_{0} \neq 0$, then in linear approximation, $f$ at a point $z = z_{0}$ is a composition of rotation by the angle $arg\, f^{\prime}(z_{0})$ and dilation by $|f^{\prime}(z_{0})|$.

So, what I think I have to do is the following: 1) Take the derivative of $f(z) = w = z^{2}$, 2) For each of (a), (b), and (c) above, evaluate $f^{\prime}(z)$ at the given value of $z_{0}$, 3) Find the argument of each of those derivatives, and 4) Take the magnitude of each derivative.

Step 3) should give me the angle I want, and Step 4) should give me the value of magnification. Is this correct?

If so, could somebody please verify if the following are correct, and if not, please tell me what I have to do to fix:

First, if $f(z) = z^{2}$, then $f^{\prime}(z) = 2z$.

(a) $f^{\prime}(z_{0})=f^{\prime}(i) = 2i$, so $\varphi \in arg \,(2i) = \frac{\pi}{2} + 2\pi n$ is an angle by which the tangents are rotated. And, the corresponding value of magnification is $|2i|=\sqrt{0^{2} + 2^{2}} = 2$.

(b) $f^{\prime}(z_{0})=f^{\prime}\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right) = -\frac{1}{2}$, so $\varphi \in arg\,\left(-\frac{1}{2}\right) = \pi + 2\pi n$ is an angle by which the tangents are rotated. And, the corresponding value of magnification is $\left\vert -\frac{1}{2} \right\vert = \frac{1}{2}$.

(c) $f^{\prime}(z_{0}) = f^{\prime}(1+i) = 2(1+i) = 2+2i$, so $\varphi \in arg\, (2(1+i)) = arg \,(1+i) = \frac{\pi}{4} + 2 \pi n$ is an angle by which the tangents are rotated. And, the corresponding value of magnification is $|2+2i|=\sqrt{2^{2} + 2^{2}} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.

It just seems all too easy, so I thought I'd check with another set of more experienced eyes!