Finding the mistake in $f(t) = (\cos (t), \sin (t)), f'(t) = if(t)$

Question

I need to undertsand what is my mistake here.

$ f(t) = (\cos(t),\sin (t))$

$ f'(t) = (-\sin(t),\cos (t))$

The tangent of a circle is orthogonal to the radius. Since orthogonality in $\mathbb{C}$ is multiplication by $i$, then:

$f'(t) = if(t)$

$ (-\sin(t),\cos (t)) = (i\cos(t),i\sin (t))$

But then $\displaystyle i=-\frac{\sin(t)}{\cos(t)} = \frac{\cos(t)}{\sin(t)} \Rightarrow 1=0$, which is absurd.

Answer 1

If you want to use orthogonality in $\mathbb{C}$, then $f(t)$ and $f'(t)$ should be considered as complex numbers, i.e.,

$f(t)=\cos(t)+i\sin(t)$

and

$f'(t)=-\sin(t)+i\cos(t)$.

Then,

$if(t)=i\cos(t)-\sin(t)=f'(t)$.

Answer 2

You can't just multiply $(\cos t,\sin t)$ by $i$. The effect is different than a rotation by $\frac {\pi}2$. To get that, add $\frac {\pi}2$to $t$...