Solutions to sin(z) = ... In Complex Numbers

[Idealistic]

Find all solutions to sin(z) = root2/2 + i*root2/2

I have a couple of methods, both of which lead to a dead end.

Method one:

sin(z) = sin(x + iy) = sin(x)cosh(y) + i*cos(x)sinh(y) = root2/2 + i*root2/2

sin(x)cosh(y) = root2/2 (Real Part)
and
cos(x)sinh(y) = root2/2 (imaginary part)

Im not sure how to solve this one without implicitly solving for x or y - It would get quite ugly.

Method 2, uses the formula:

e[sup:1lmgk6ek]iz[/sup:1lmgk6ek] = i*(w) + (1 - w[sup:1lmgk6ek]2[/sup:1lmgk6ek])[sup:1lmgk6ek]1/2[/sup:1lmgk6ek] (where w = sin(z), in my case w = root2/2 + i*root2/2)

so

e[sup:1lmgk6ek]iz[/sup:1lmgk6ek] = i(root2/2 + i*root2/2) + (1 - ((root2/2) + i(root2/2))[sup:1lmgk6ek]2[/sup:1lmgk6ek])[sup:1lmgk6ek]1/2[/sup:1lmgk6ek]

e[sup:1lmgk6ek]iz[/sup:1lmgk6ek] = i(root2/2 + i*root2/2) + (1 - i)[sup:1lmgk6ek]1/2[/sup:1lmgk6ek]

if I find the second roots of 1 - i, i get:

2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek](cos(-pi/8) + i*sin(-pi/8))
and
2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek](cos(7pi/8) + i*sin(7pi/8))

then I use those roots and put them in to this equation,

z =
-i*log(-root2/2 + 2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek]cos(-pi/8) + i(root2/2 + 2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek]sin(-pi/8))
and with the other root
-i*log(-root2/2 + 2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek]cos(7pi/8) + i(root2/2 + 2[sup:1lmgk6ek]1/4[/sup:1lmgk6ek]sin(7pi/8))

...

With either method, the first of which seems smoother, I i'm having trouble with finding a solution. Can someone guide me in the right direction?

.

 

[soroban]

Hello, Idealistic!

I tried something different . . . please check my work.

\(\displaystyle \text{Find all solutions to: }\; \sin z \:=\:\tfrac{\sqrt{2}}{2} + i\tfrac{\sqrt{2}}{2}\)

\(\displaystyle \text{I used the complex definition of }\sin z\!:\;\;\sin z \:=\:\frac{e^{iz} - e^{-iz}}{2i}\)

\(\displaystyle \text{We have: }\:\frac{e^{iz} - e^{-iz}}{2i} \;=\;\frac{\sqrt{2}}{2}(1+i)\)


\(\displaystyle \text{Multiply by }2i\!:\;\;e^{iz} - e^{-iz} \;=\;i\sqrt{2}(1+i) \;=\;-\sqrt{2}((1-i)\)


\(\displaystyle \text{Multiply by }e^{iz}\!:\;\;e^{2iz} - 1 \;=\;-\sqrt{2}(1-i)e^{iz} \quad\Rightarrow\quad e^{2iz} + \sqrt{2}(1-i)e^{iz} - 1 \:=\:0\)


\(\displaystyle \text{Quadratic Formula: }\;e^{iz} \;=\;\frac{-\sqrt{2}(1-i) \pm \sqrt{2(1-i)^2 + 4}}{2} \;=\;\frac{-\sqrt{2}(1-i) \pm2\sqrt{1-i}}{2}\)


\(\displaystyle \text{Take logs: }\;\ln\left(e^{iz}\right) \;=\;\ln\left[\frac{-\sqrt{2}(1-i) \pm 2\sqrt{1-i}}{2}\right]\)

. . . . . . . . \(\displaystyle iz\ln(e) \;=\;\ln\left[\frac{-\sqrt{2}(1-i) \pm \sqrt{1-i}}{2}\right]\)

. . . . . . . . . . . .\(\displaystyle iz \;=\;\ln\left[\frac{-\sqrt{2}(1-i) \pm \sqrt{1-i}}{2}\right]\)

. . . . . . . . . . . . \(\displaystyle z \;=\;\frac{1}{i}\ln\left[\frac{-\sqrt{2}(1-i) \pm\sqrt{1-i}}{2}\right]\)


Can this be simplified? . . . I have no idea!