Question

I need find z such that tan z = i/2 . I post my attempt and where I stuck. Please, help me out . BTW, the site is on and off at my end. Hence, if you don't see me, it doesn't mean I don't care your opinion. Thanks in advance.

Answer 1

I get \[\frac{\ln(3)}{2}i\]

solve

\[\frac{e^{\theta}-e^{-\theta}}{e^{\theta}+e^{-\theta}}=\frac{1}{2}~\text{ ,for }~\theta\]

Answer 2

I got it. Thank you.

Answer 3

My solution
\(tan z = \dfrac{sinz}{cos z}= \dfrac{\frac{e^{iz}-e^{-iz}}{2i}}{\dfrac{e^{iz}+e^{-iz}}{2}} =\dfrac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=\dfrac{i}{2}\)

Answer 4

@freckles watch me, please

Answer 5

\(e^{iz}-e^{-iz} = -\dfrac{1}{2}(e^{iz}+e^{-iz})\)

Answer 6

\(2e^{iz}-2e^{-iz}= -e^{iz} -e^{-iz}\)

Answer 7

\(3e^{iz} -e^{-iz}=0\)
multiple both sides by \(e^{iz}\)
we have
\(3e^{2iz} -1=0\\e^{2iz}= \dfrac{1}{3}\)

Answer 8

agree with everything you have so far

Answer 9

\(\implies 2iz = ln (1/3) \\z = \dfrac{ln(1/3)}{2i} = -i\dfrac{ln (1/3)}{2} = i\dfrac{ln 3}{2}\)

Answer 10

It is matching with what Zarkon gave me above

Answer 11

And next is my classmate solution.

Answer 12

should you write all the solutions...
i*ln(3)/2-n*pi

Answer 13

how?

Answer 14

\[e^{2 i z}=e^{(2z)i}=e^{(2z+2 n \pi) i} \\ e^{(2 z + 2 n \pi )i}=\frac{1}{3} \\ (2 z + 2 n \pi) i =\ln(\frac{1}{3}) \\ 2 i (z + n \pi)=\ln(\frac{1}{3}) \\ 2i^2 (z+n \pi )=i \ln(\frac{1}{3} )\\ z+n \pi =\frac{-1}{2} i \ln (\frac{1}{3}) \\ z + n \pi =i \frac{1}{2}\ln((\frac{1}{3})^{(-1)}) \\ \\ z+n \pi =\frac{i \ln(3) }{2}\\ z=\frac{i \ln(3)}{2}- n \pi\]

Answer 15

Ok, now see what my classmate did.
From \(e^{iz}-e^{-iz} = -\dfrac{1}{2}(e^{iz}+e^{-iz}) \)
\(e^{iz}-e^{-iz} = -\dfrac{1}{2}(e^{iz})-\dfrac{1}{2}(e^{-iz})\)
Let \(x = e^{iz}\)
He got \(x - \dfrac{1}{x} = -\dfrac{1}{2}x - \dfrac{1}{2x}\)

Answer 16

That leads to the equation I asked and the solution is \(x =\pm\dfrac{\sqrt3}{3}\)

Answer 17

When plug back, he got \(e^{iz}= \pm\dfrac{\sqrt 3}{3}\)
You see, it different from what we do above

Answer 18

\[e^{iz}= \frac{\sqrt{3}}{3} \\ e^{(z+2 n \pi)i}=\frac{\sqrt{3}}{3} \\ (z+2 n \pi)i=\ln(\frac{\sqrt{3}}{3}) \\ z+2n \pi =-i \ln(\frac{\sqrt{3}}{3}) \\ z+2 n \pi =i \ln( \frac{3}{\sqrt{3}}) \\ z+ 2 n \pi =i \ln(\sqrt{3}) \\ z=i \ln (\sqrt{3})-2 n \pi \\ \text{ and similarly solving } \\ e^{iz } =-\frac{\sqrt{3}}{3} \\ \text{ we get } \\ z=i \ln(-\sqrt{3})- 2 n \pi \\ \text{ but } \ln(-\sqrt{3})=\ln(3)+i \pi \\ \text{ so we have } \\ z=i \ln(\sqrt{3})+i \pi -2 n \pi \\ z=i \ln(\sqrt{3})-n \pi \]

Answer 19

the second solution there actually contains the first solution

Answer 20

one sec I made a mistake

Answer 21

\[\ln(-\sqrt{3})=\ln(\sqrt{3})+i \pi \text{ first I made a typeo here } \\ \text{ second I forgot \to do the following } \\ z=i \ln (-\sqrt{3})-2n \pi \\ z=i(\ln(\sqrt{3})+i \pi)-2n \pi \\ z=i \ln (\sqrt{3})- \pi -2n \pi \\ z=i \ln(\sqrt{3})-\pi (1+2n)\]

Answer 22

but anyways that solution gives us:
\[z=i \ln(\sqrt{3})-\pi(1+2 n) \text{ or } z=i \ln(\sqrt{3})-\pi (2n)\]
if I didn't do something wrong

Answer 23

and of course you can write i ln(sqrt(3)) as i/2 *ln(3)

Answer 24

that last thing I wrote is your solution

and I believe the solutions are equivalent yours and your friend's

Answer 25

Thank you so much. We discuss but can't find what is wrong on both solutions.

Answer 26

is there suppose to be something wrong?
is \[i \frac{1}{2} \ln(3)- n \pi \text{ not correct ?}\]
or the other way to write it ... \[i \frac{1}{2} \ln(3)-2n \pi \text{ or } i \frac{1}{2}\ln(3)-(2n+1) \pi\]?

Answer 27

like I guess what I'm asking
are you still seeing differences in the final solution ?

Answer 28

oh, I know how to Not redo my solution but still get Your answer
From \(z = \dfrac{i}{2} ln 3\)
we have \( |3|= 3, arg 3 =0 + 2k\pi\)
hence \(ln 3 = ln 3 + i (2k\pi)\)
Then
\(z = \dfrac{i}{2} (ln 3 + i 2k\pi) = \dfrac{iln3}{2} - k\pi\)

Answer 29

ok cool stuff

Answer 30

:)

Answer 31

Again, thank you so much. Have a good night.

Answer 32

you too