1. ## sin(z) = cos(z)

$\displaystyle z\in\mathbb{C}$

$\displaystyle sin(z)=cos(z)\Rightarrow sin(z)-cos(z)=0$

$\displaystyle \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}-\frac{e^{\mathbf{i}z}+e^{-\mathbf{i}z}}{2}=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}-e^{-\mathbf{i}z}-\mathbf{i}e^{\mathbf{i}z}-\mathbf{i}e^{-\mathbf{i}z}=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}(1-\mathbf{i})=e^{-\mathbf{i}z}(1+\mathbf{i})$

$\displaystyle \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}=1$

$\displaystyle \displaystyle -\mathbf{i}e^{2\mathbf{i}z}=1\Rightarrow e^{2\mathbf{i}z}=\mathbf{i}$

Book's answer $\displaystyle \displaystyle z=\frac{\pi}{4}(4k+1) \ k\in\mathbb{Z}$.

Thanks.

2. Originally Posted by dwsmith
$\displaystyle z\in\mathbb{C}$

$\displaystyle sin(z)=cos(z)\Rightarrow sin(z)-cos(z)=0$

$\displaystyle \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}-\frac{e^{\mathbf{i}z}+e^{-\mathbf{i}z}}{2}=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}-e^{-\mathbf{i}z}-\mathbf{i}e^{\mathbf{i}z}-\mathbf{i}e^{-\mathbf{i}z}=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}(1-\mathbf{i})=e^{-\mathbf{i}z}(1+\mathbf{i})$

$\displaystyle \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}=0$ Mr F says: This line should be $\displaystyle \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}$=1.

[snip]
..

3. Picking up from the correction:

$\displaystyle \displaystyle e^{\mathbf{i}z}=\pm\sqrt{\mathbf{i}}\Rightarrow \mathbf{i}z=ln(\pm\sqrt{\mathbf{i}})$

$\displaystyle \displaystyle \Rightarrow z=-\mathbf{i}\left(ln|\sqrt{\mathbf{i}}|+\mathbf{i}\l eft(\frac{\pm\pi}{2}+2\pi k\right)\right) \ k\in\mathbb{Z}$

$\displaystyle \displaystyle z=\frac{\pm\pi}{2}+2\pi k=\pi \left(\frac{\pm 1}{2}+k\right) \ \ k\in\mathbb{Z}$

Something is still wrong though....

The arg of z is negative and positive $\displaystyle \displaystyle \frac{\pi}{2}+2\pi k$ and modulus of z is positive and negative i.

4. You are solving $\displaystyle \exp(2zi)=i$.
That means $\displaystyle e^{-2y}\cos(2x)=0~\&~ e^{-2y}\sin(2x)=1$.
That gives $\displaystyle 2y=0~&~2x=\frac{\pi}{2}+2k\pi$ or $\displaystyle x=\frac{\pi}{4}+k\pi.$
That is the answer given the by the text.

5. Hello, dwsmith!

$\displaystyle \sin z\,=\,\cos z, \;\;z\in \mathbb{C}$

Since neither $\displaystyle \dfrac{\pi}{2}$ nor $\displaystyle \dfrac{3\pi}{2}$ are roots of the equation, $\displaystyle \cos z \ne 0$
Divide by $\displaystyle \cos z\!:\;\;\dfrac{\sin z}{\cos z} \:=\:1 \quad\Rightarrow\quad \tan z \:=\:1$
Therefore: .$\displaystyle z \;=\;\frac{\pi}{4} + \pi k$