Statement
$(i)$ Let $\gamma:\mathbb R \to \mathbb C$ a $C^1$ curve. Let $v={\gamma}'(t_0)$ the complex number that one obtains from translating to the origin the tangent vector to $\gamma$ at $t=t_0$. Let $f:\mathbb C \to \mathbb C$ be holomorphic and let $z=f'(\gamma(t_0))$. Prove that $zv$ is the complex number obtained from translating to the origin the tangent vector of the curve $f \circ \gamma$ at $t=t_0$.
$(ii)$Let $\gamma_1,\gamma_2:\mathbb R \to \mathbb C$ defined by $\gamma_1(t)=t$ and $\gamma_2(t)=(1+i)t$. Let $f:\mathbb C \to \mathbb C, f(z)=\sin(z)+z^4$. Calculate at which angle the curves $f \circ \gamma_1$ and $f \circ \gamma_2$ intersect at $0$.
My attempt
I don't know what to do for $(i)$.
Now, for $(ii)$, by simple calculation $$f \circ \gamma_1 (0)=(\sin(t)+t^4) |_{t=0}=0$$ and $$f \circ \gamma_2 (0) =(\sin((1+i)t)+((1+i)t)^4)|_{t=0}=0$$
So this verifies that $f \circ \gamma_1$ and $f \circ \gamma_2$ intersect at $t=0$.
To know the angle of intersection, I suspect that I have to calculate the angle between the tangent vectors of $f \circ \gamma_1$ and $f \circ \gamma_2$ respectively at $t=0$. So these tangent vectors are $f'(\gamma_1(0)){\gamma_1}'(0)$ and $f'(\gamma_2(0)){\gamma_2}'(0)$. Using $(i)$ and by direct calculation, I obtain $$f'(\gamma_1(0)){\gamma_1}'(0)=1$$ $$f'(\gamma_2(0)){\gamma_2}'(0)=1+i$$
So the angle is $\arccos(\dfrac{Re(1.1+0.i)}{\sqrt{2}})=\dfrac{\pi}{4}$, or $45$ degrees.
I would like some help to show part $(i)$ and I would like to know if what I've done in $(ii)$ is correct.