That turned dark quickly. Instead I wrote: $$\sin(\alpha\pi) \text{i} - \cos(\alpha\pi)\pi i = 0\tag{4}$$ That left the original equation (2) unsatisfied and the book then said: "The general solution is $z_1 = c_1 \exp(\alpha\pi) \text{i}$ $z_2 = c_2 \exp(\alpha\pi) \text{i}$" $c_1, c_2 \in \mathbb{R}$ As I was not able to find any mistake in my workings, I'm not sure what I missed here. Any idea? A: The question did not mention $\cos \alpha \pi$ so the answer should be $\sin\alpha \pi$ A: You seem to be confused about what exactly is going on. If you have an expression like $z=x+iy$, then you can certainly take its cosine and sine separately, and for any numbers, cos(x+y) and sin(x+y). Now, there are things that aren't true, such as $\sin(x\pm y)$ not being equal to $\sin x\cos y \pm \cos x \sin y$. Another thing that's not true is that $\sin (x+y)=\sin x \cos y + \cos x \sin y$, even if you use only real values for x and y. So what can you do? Well, $\sin(x+y)$ is equal to $\sin x \cos y + \cos x \sin y$, but only when x and y are between $0$ and $2\pi$ and $x+y \in [0,2\pi]$. That is, your sine function is only defined for values of x and y where both sin and cos are between $-1$ and $1$, and such that the sum is within the same range, i.e. between $0$ and $2\pi$. A: If $z=\cos(\alpha \pi)+\text{i} \sin(\alpha \pi) , \ \text{i}^2=-1$ and $ \ \text{i} z= \sin(\alpha \pi)+ \cos(\alpha \pi) \text{i}=0$ which can happen only for $z=0$, we conclude that $\cos(\alpha \pi)=0$ which gives $\alpha \pi=2k \pi$ or $ \alpha= \frac{2k \pi}{ \pi}$ which is not the case. A: If $$z = cos(\alpha \pi) + \text{i} sin(\alpha \pi) = 0$$ Then $z = z_1 + z_2 \implies \cos(\alpha \pi) = 0$ Which gives you two real solutions for $\alpha \pi$ , viz. $0$ and $\pi$. $\text{i} \text{s}in(\alpha \pi) = \text{i} 0 + \text{i} \pi \implies \pi = 2k \pi \text{ or } 2k \pi (1-\text{i})$ $$\implies \alpha = \pi / 2k \text{ or } \pi/2k (1+\text{i})$$ Which are clearly not integers. Edit : Or use the double angle formula $\cos 2 \theta = 2 \cos^2 \theta - 1$, $$\implies z = \cos(\alpha \pi) + \text{i} \sin(\alpha \pi) = 0 \implies \cos(\alpha \pi) = 0 \implies \alpha = \frac{2k \pi}{ \pi} \text{ or } \frac{2k \pi}{ \pi} (1-\text{i}) \implies \alpha = \frac{\pi}{2k \text{ or } 2k} $$ As for the given answer. There is no issue of modulus $\pm 1$ here. $$\text{If} \ \alpha = \frac{\pi}{2k} \ \text{ and }\ 1-\text{i} \ \text{then } \frac{1-\text{i}}{1+\text{i}} = -\frac{1}{2} \in \mathbb{R}$$ $$\text{If} \ \alpha = \frac{\pi}{2k} \ \text{ and } \ 1+\text{i} \ \text{then } \frac{1+\text{i}}{1-\text{i}} = \frac{1}{2} \in \mathbb{R}$$ The denominator $1 \pm \text{i}$ is always $1 \mp \text{i}$ for integer $k$. In fact, I've found the source of this answer. Reference - Page $28$ of the book : "Essential Problem Solving Using Trigonometry" by H.F. Umar It looks like the book contains typographical error(s) and/or a misunderstanding/misprint. In case the book continues to say what it says. :P Here is the complete answer of the question. The answer for $\alpha$ is given by : $\text{If} \ \alpha = \frac{2k \pi}{ \pi} \ \text{ then } \alpha = \frac{\pi}{2k}$ $\text{If} \ \alpha = \frac{2k \pi}{ \pi} \ \text{ then } \alpha = \frac{\pi}{2k} \text{ where } k \in \mathbb{Z} \setminus \{0\}$ As mentioned in the beginning of the post, $\text{if } \ \alpha \text{ is an integer then } \cos \alpha \text{ and } \sin \alpha \text{ are always of modulus }\pm 1$ To evaluate the modulus of $\alpha$ by $\cos \alpha$ or $\sin \alpha$,$\alpha$ must be an integer. So the answer for $z