Is $\csc x = a+bi$ defined?

Question

A question in my assignment asks to find which equation can have its roots $\sec^2 x, \csc^2 x$ (no restrictions in $x$ were given).

The equations were quadratic ($x^2+bx+c$) and the one which satisfies $-b=c$ can be the required equations.

There were two which satisfies this,

$x^2-3x+3=0, x^2-9x+9=0$

The second one is absolutely correct but the first in the first one $\csc^2x$ is in the form of $a+bi$. And so first one wasn't the correct answer.

But I am having doubt in the answer, as by Euler's formula,

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

And here putting a suitable complex value of $x$, $\sin x$ can have a value in form of $a+bi$ and so $\csc^2 x$.

So, am I correct? Can we put a complex value in trigonometric functions or can they have a complex value? And also tell what if $\sin x=2$?

Answer 1

Yes, but your notation is a bit unconventional. People usually write $z=x+yi$ where $x,y \in \mathbb R$. So, writing Euler's formulae for $\sin$ and $\cos$, we have sensible definitions for all the trigonometric functions evaluated at a complex number and these definitions, when restricted to the real numbers, agree with the real trigonometric functions. Moreover, the addition formulae for $\sin(z_1+z_2)$ and $\cos(z_1+z_2)$ as well as such old favourites as $\sin^2(z)+\cos^2(z)=1$ work exactly as one would hope. To solve $\sin(z)=2$, equate real andimaginary parts.

Answer 2

If you want to solve the equation $$\csc (x+i y)=a+i\,b$$ expand first the cosecant, go to hyperbolic functions, use the conjugate, separate the real and imaginary parts to end with $$\frac{\sin (x) \cosh (y)}{\cos ^2(x) \sinh ^2(y)+\sin ^2(x) \cosh ^2(y)}=a \tag 1$$ $$\frac{\cos (x) \sinh (y)}{\cos ^2(x) \sinh ^2(y)+\sin ^2(x) \cosh ^2(y)}=-b \tag 2$$ Computing the ratio $$\tan (x) \coth (y)=-\frac ab\quad \implies \quad \color{blue}{y=-\coth ^{-1}\left(\frac{a }{b}\cot (x)\right)} \tag 3$$ Replacing $y$ in $(1)$ give the simple $$\csc (x) \sqrt{1-\frac{b^2 }{a^2}\tan ^2(x)}=\frac{a^2+b^2}{a}$$ Squaring $$\csc ^2(x)-\frac{b^2 }{a^2}\sec ^2(x)=\left(\frac{a^2+b^2}{a}\right)^2$$ Cross multiply by $\sin^2(x)\,\cos^2(x)$ $$\cos ^2(x)-\frac{b^2 }{a^2}\sin ^2(x)=\left(\frac{a^2+b^2}{a}\right)^2\sin^2(x)\,\cos^2(x)$$ Let $x=\sin^{-1}(t)$ to obtain $$t^4-\frac{a^2+b^2+1}{a^2+b^2} t^2+\frac{a^2}{\left(a^2+b^2\right)^2}=0$$ which is just a quadratic in $t^2$.