1
$\begingroup$

I tried solving the equation $\cos{z}=2i$ where $z$ is a complex number. The solution $i$ have ended up with is $z$, $= (4k+1)\frac{\pi}{2} - i\ln{(2+\sqrt{5})}$. However the text book solution is $z=(2k+1)\frac{\pi}{2} - (-1)^k\ln{(2+\sqrt{5})}i$. Are the results geometrically same or my solution is erroneous.

$\endgroup$
  • 2
    $\begingroup$ It would help if you showed how you got your solution. $\endgroup$ – Thomas Andrews Oct 5 '15 at 14:06
  • $\begingroup$ Related: math.stackexchange.com/q/1242873 $\endgroup$ – Did Oct 5 '15 at 14:13
  • $\begingroup$ thanx guys...Thomas and H.R. your solns are elegant...I was skipping the 2-sqrt(5) bcoz everytime i took the log it was coming log(2-sqrt(5)) and was thinking log of negative number is not possible. $\endgroup$ – rotating_image Oct 5 '15 at 15:24
5
$\begingroup$

$$\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right).$$ Solve first for $w=e^{iz}$.

$$w^2-4iw + 1=0$$ so: $$w=\frac{4i\pm\sqrt{-16-4}}{2} = (2\pm\sqrt{5})i$$

So we have to deal with two cases with care. If $w=(2+\sqrt{5})i$, you get your solution:

$$z = (4k+1)\frac{\pi}{2} - i\ln(2+\sqrt{5})$$

If $w=(2-\sqrt{5})i$, then using $2-\sqrt{5}=-(2+\sqrt{5})^{-1}$, we have:

$$iz = -\log(2+ \sqrt{5}) +\left(\frac{3\pi}{2}+2\pi k\right)i$$

So:

$$z = \left(\frac{3pi}{2}+2\pi k\right) + \log(2+ \sqrt{5})i$$

In the first case, let $m=2k$ and you have:

$$z = (2m+1)\frac{\pi/2} - (-1)^{m}\ln(2+\sqrt{5})i$$

and the second case $m=2k+1$ then:

$$ z = (2m+1)\frac{\pi}{2} - (-1)^{2m+1}\ln(2+\sqrt{5})$$

So the full answer merges the two answers.


You can probably write this carefully as one case. The two values of $w$ can be written as $$w=i(-1)^k(2+\sqrt{5})^{(-1)^k}$$. Then:

$$iz = \log w = (-1)^k \log(2+\sqrt{5}) + \frac{\pi}{2}i + k\pi i$$

So:

$$z = \left(k+1/2\right)\pi - (-1)^k \log(2+\sqrt{5}) i$$

$\endgroup$
2
$\begingroup$

Your equation (for $k$) holds, since it is the equation from your book for $\hat{k}=2k$.

If $\hat{k}$ is impair such that the equation from the book holds there can not be a $k$ such that your equation for $k$ becomes the equation from the book for $\hat{k}$ since that would imply that $-1=\left(-1\right)^\hat{k}=+1$.

(two complex numbers coincide iff their real and complex coincide, I considered the complex and devided by $-\ln\left(2+\sqrt{5}\right)$)

The conclusion is, that you found only half of the solutions.

It seems like you used $\cos\left(x\right)=\cos\left(2\pi+x\right)$ but not $\cos\left(x\right)=\cos\left(-x\right)$.

$\endgroup$
2
$\begingroup$

You found half the answers.

Hint; What I usually do to solve these wrong-proof is write

$\cos x=\frac{e^{ix}+e^{-ix}}{2}$

And let $y:=e^{ix}$

Hint#2: Quadratics can have 2 different solutions.

$\endgroup$
1
$\begingroup$

Here is the solution without explanations:

$$\eqalign{ & \cos (z) = 2i \cr & \cos (x + iy) = \cos (x)\cos (iy) - \sin (x)\sin (iy) = 2i \cr & \cos (x)\left( {{{{e^{i(iy)}} + {e^{ - i(iy)}}} \over 2}} \right) - \sin (x)\left( {{{{e^{i(iy)}} - {e^{ - i(iy)}}} \over {2i}}} \right) = 2i \cr & \cos (x)\left( {{{{e^{ - y}} + {e^y}} \over 2}} \right) - \sin (x)\left( {{{{e^{ - y}} - {e^y}} \over {2i}}} \right) = 2i \cr & \cosh (y)\cos (x) + {1 \over i}\sinh (y)\sin (x) = 2i \cr & \cosh (y)\cos (x) - i\sinh (y)\sin (x) = 2i \cr & \left\{ \matrix{ \cosh (y)\cos (x) = 0 \hfill \cr \sinh (y)\sin (x) = - 2 \hfill \cr} \right. \cr & \left\{ \matrix{ \cos (x) = 0 \to x = (2n + 1){\pi \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n = 0,1,2,... \hfill \cr \left\{ \matrix{ \sinh (y)\sin (x) = \sinh (y){( - 1)^n} = - 2 \to \sinh (y){( - 1)^{n + 1}} = 2 \to \sinh ({( - 1)^{n + 1}}y) = 2 \hfill \cr {( - 1)^{n + 1}}y = {\sinh ^{ - 1}}(2) = \ln (2 + \sqrt 5 ) \to y = {( - 1)^{n + 1}}\ln (2 + \sqrt 5 ) \hfill \cr} \right. \hfill \cr} \right. \cr & z = (2n + 1){\pi \over 2} + i{( - 1)^{n + 1}}\ln (2 + \sqrt 5 ) \cr} $$

In final steps, I used the fact that $\sinh(y)$ is an odd function of $y$.

$\endgroup$
  • $\begingroup$ @NimaBavari: Thanks dude. :) $\endgroup$ – Hosein Rahnama Feb 13 '16 at 8:57
  • $\begingroup$ Khahesh mikonam. $\endgroup$ – user98186 Feb 13 '16 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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