Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To find the complex numbers z satisfying $\sin(z) = \cos(z)$, can I say: $$\sin(z) = \frac{(e^{iz}-e^{-iz})}{2i}=\frac{(e^{iz}+e^{-iz})}{2}$$

and solve for z? So we then reduce this to $$-e^{-iz} = e^{-iz}$$ but this doesn't look right

share|improve this question
    
I don't see how you get from the first equation to the second one. –  celtschk Jul 13 '12 at 6:23
add comment

3 Answers 3

Use angle addition identities and the relations between trig and hyperbolic trig functions.

$$\begin{align} &&\sin(z)&=\cos(z)\\ \implies&&\sin(a+bi)&=\cos(a+bi)\\ \implies&&\sin(a)\cos(bi)+\cos(a)\sin(bi)&=\cos(a)\cos(bi)-\sin(a)\sin(bi)\\ \implies&&\sin(a)\cosh(b)+i\cos(a)\sinh(b)&=\cos(a)\cosh(b)-i\sin(a)\sinh(b)\\ \end{align}$$

Now equating real parts (and keeping in mind that $a$ and $b$ are real), we see that $\sin(a)=\cos(a)$. (This also use the fact that $\cosh(b)$ cannot be $0$.) So $a=\frac{\pi}{4}+k\pi$ for some $k\in\mathbb{Z}$.

Now we can divide across by $\sin(a)$ or $\cos(a)$. (Remember, we have deduced they are equal.)
$$\begin{align} \implies&&\cosh(b)+i\sinh(b)&=\cosh(b)-i\sinh(b)\\ \implies&&i\sinh(b)&=-i\sinh(b)\\ \implies&&\sinh(b)&=-\sinh(b)\\ \implies&&\sinh(b)&=0\\ \end{align}$$ There is only one real solution for $b$: $b=0$.

So in conclusion, $z=\frac{\pi}{4}+k\pi$ for some $k\in\mathbb{Z}$.

share|improve this answer
    
This to me is the neatest route. +1, of course. –  J. M. Jul 13 '12 at 7:43
add comment

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$ Then, $$\sin(z)=\cos(z)\implies \frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}$$ Now let $e^{iz}=t$ ,then $e^{-iz}=1/t$ and you will get a quadratic equation,solve for it and back substitute it to get $z$.

The equation becomes $t^2=\frac{1+i}{1-i}=i=e^{i(\pi/2+2k\pi)}\implies t=e^{i(\pi/4+k\pi)},e^{i(5\pi/4+k\pi)}$. Equating it with $e^{iz}$ we get $z=\pi/4+k\pi$

share|improve this answer
    
so I get 2+i/(2-i) = t^2 and t = +/- sqrt(3/5+4/5i)?? –  mary Jul 13 '12 at 6:03
    
I think you should get $t^2=\frac{1+i}{1-i}$. –  Aang Jul 13 '12 at 6:10
    
The last sentence should read: we get $z=\pi/4+k\pi$ for some $k$ in $\mathbb Z$. –  Did Jul 13 '12 at 6:24
    
right, that would be the general solution. –  Aang Jul 13 '12 at 6:25
add comment

The definition of $\sin(z)$ is $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.