In school I am learning for complex numbers in trigonometric form.
$$z= a + bi r ( \cos(\alpha) + i\sin(\alpha) )$$
In a problem I have to find the real and imaginary part from trigonometric form of
$$1 + \cos(\alpha) + i\sin(\alpha)$$
For which I think the solution is
- Real part = $1+ \arccos(\alpha)$
- Imaginary part = $\arcsin(\alpha)$
PR2: $\sin(\alpha)+i\cos(\alpha)$
Update>
For which I think the solution is>
- Real part = $\arcsin(\alpha)$
- Imaginary part = $\arccos(\alpha)$
Best Answer
HINT
Note if you have a trigonometric form $$z = r(\cos a + i \sin a) = r\cos a + i r\sin a,$$ the imaginary part is $r\sin a$ and the real part is $r \cos a$.
E.g. if $z = 1 + i\sin(\pi/7) + \cos(\pi/13)$ then the imaginary part is the piece which is the multiple of $i$, so it is $\sin(\pi/7)$ and everything else is the real part, which here is $1 + \cos(\pi/13)$.
Another one to make it clear. Let $$ z = 2 + 3i + \cos(\pi/5)i + \pi = (2+\pi) + i \left[3 + \cos(\pi/5) \right], $$ So the real part is $2+\pi$ and the imaginary part is $3 + \cos(\pi/5)$.