$\cos(iz)=\sin(iz)$

Question

$$\cos(iz)=\sin(iz),\quad z\in\Bbb C$$ How do we solve this? I started replacing $z=x+iy$ but can't continue. Any suggestions?

Answer 1

I think you treat this question too complicated.

Let $x=iz$ ,

Then $\cos x=\sin x$

$\cos x=\cos\left(\dfrac{\pi}{2}-x\right)$

$x=2n\pi+\dfrac{\pi}{2}-x$ , $\forall n\in\mathbb{Z}$

$2x=\dfrac{(4n+1)\pi}{2}$ , $\forall n\in\mathbb{Z}$

$x=\dfrac{(4n+1)\pi}{4}$ , $\forall n\in\mathbb{Z}$

$iz=\dfrac{(4n+1)\pi}{4}$ , $\forall n\in\mathbb{Z}$

$z=-\dfrac{(4n+1)\pi i}{4}$ , $\forall n\in\mathbb{Z}$

Answer 2

So, $\tan(iz)=1$

So, $iz=m\pi+\frac\pi 4$ where $m$ is any integer.

$z=\frac 1 i(m\pi+\frac\pi 4)=-i\frac{(4m+1)\pi }4$

Now, the difference of two consecutive values of $iz$ is $-i\frac{\{4(r)+1\}\pi }4-\left(-i\frac{\{4(r+1)+1\}\pi }4\right)=\pi i$

As we know $e^{2\pi i}=1$ and $$\tan(iz)=1\iff \frac{e^z+e^{-z}}{e^z-e^{-z}}=i\iff e^{2z}=-i$$ there will be $\frac{2\pi}\pi=2$ distinct solutions(Compare the result with that of the next method).

---

Using Euler's identity and Euler's formula,

$$\cos (iz)=\frac{e^{i(iz)}+e^{-i(iz)}}2=\frac{e^z+e^{-z}}2$$

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

So, $$\frac{e^z+e^{-z}}2=i\frac{e^z-e^{-z}}2$$

or, $$e^{2z}+1=i(e^{2z}-1)\implies e^{2z}=-i=e^{-\frac{i\pi}2}$$ for $-i=R(\cos A+i\sin A)\implies R=\sqrt{0^2+(-1)^2}=1$ and $\cos A=0,\sin A =-1\implies A=-\frac{i\pi}2$

So, $$ e^{2z}=e^{2n\pi i-\frac{i\pi}2}$$ where $n$ is any integer as $e^{2n\pi i}=\cos(2n\pi)+i(\sin2n\pi)=1$.

So, $z=\frac 12 (2n\pi i-\frac{i\pi}2)=i\frac\pi 4(4n-1)$ where $0\le n<2$ or more generally any two in-congruent values of $n\pmod 2$ will give us the 2 distinct solution of $e^{2z}=-i$ and consequently, of the given equation.

Answer 3

What if you write $\cos(z)=\frac 12 (e^{iz}+e^{-iz})$ and $\sin(z)=\frac 1{2i}(e^{iz}-e^{-iz})$?

Answer 4

Putting, as you did, $\,z=x+iy\,$ , but only in a later step, we get:

$$\frac{e^{-z}+e^z}{2}=:\cos iz=\sin iz:=\frac{e^{-z}-e^z}{2i}\Longrightarrow$$

$$ie^{-z}+ie^z=e^{-z}-e^z\Longrightarrow i(1+e^{2z})=1-e^{2z}\Longrightarrow$$

$$i[1+e^{2x}(\cos 2y+i\sin 2y)]=1-e^{2x}[\cos 2y+i\sin 2y)]\Longrightarrow$$

$$\left(-e^{2x}\sin 2y\right)+i[1+e^{2x}\cos 2y]=\left(1-e^{2x}\cos 2y\right)-ie^{2x}\sin 2y$$

Comparing real and imaginary parts:

$$-e^{2x}\sin 2y=1-e^{2x}\cos 2y$$

$$1+e^{2x}\cos 2y=-e^{2x}\sin 2y$$

Can you take it from here?

Answer 5

Let $$w=iz.$$ Then $$\cos w=\sin w,$$ $$1=\cos^2w+\sin^2w=2\cos^2w=2\sin^2w,$$ $$\cos^2w=\sin^2w=\frac12,$$ $$\cos w=\sin w=\frac{\pm1}{\sqrt2},$$ $$w=\frac\pi4+n\pi,\ n\in\mathbb Z,$$ $$z=\frac wi=-iw=-\left(n+\frac14\right)\pi i,\ n\in\mathbb Z.$$