Question

a)Prove that the equation tan z =z has only real roots. where z is complex
b)Evaluate tanh^-1(infinity)

Answer 1

@KingGeorge

Answer 2

a)i think z=x+iy since is a complex variable so tan z = tan (x+iy) so how to find the roots

Answer 3

I'm not sure if this will work, but what I might do, is write\[\tan(x+iy)=\frac{\tan(x)+\tan(iy)}{1-\tan(x)\tan(iy)}=x+iy\]and see if that gets you anywhere. Although this might just be making it more difficult.

Answer 4

actually how to find the roots of trig functions

Answer 5

for finding the roots for normal complax number i know it is
\[z=r ^{\frac{ 1 }{ n }}\left[ cis \frac{ 2\pi k +\theta }{ n } \right] \] for k=0,1,2,...

Answer 6

Well, you could also try using the formula\[\tan(z)=\frac{ie^{-iz}-ie^{iz}}{e^{-iz}+e^{iz}}\]

Answer 7

so it will be \[= \frac{ icis(-z)-icis(z) }{ cis(-z)+cis(z) }\]

Answer 8

I suppose that's another way to write it. But again, I'm not sure if that's the best way. This problem kind of has me stumped right now.

Answer 9

it is sort of 0/0

Answer 10

oops i mean 0/(2cosz) which is zero

Answer 11

Alright, I did some searching, and found a solution that makes sense.

First, write \(\tan(x+iy)\) as\[\tan(x+iy)=\frac{\sin2x+i\sinh(2y)}{\cos2x+\cosh2y}=x+iy.\]Now, all the variables and functions used are real-valued, and when viewed as vectors, we can see that \((x,y)\) and \((\sin(2x),\sinh(2y))\) are proportional to each other. Thus, the determinant must be 0.\[\begin{array}{|cc|}x & y \\ \sin2x &\sinh2y \end{array}=0.\]Finally, you apply the inequalities \(|\sin(x)|\le|x|\) and \(|\sinh(x)|\ge|x|\), which have equality if and only if \(x=0\).

Source: http://math.stackexchange.com/questions/388673/prove-that-the-equation-tan-z-z-has-only-real-roots

Answer 12

So you get \[x\sinh(2y)=y\sin(2x).\]Applying the two inequalities, we get that \[|x\sinh(2y)|\ge|2xy|\]\[|y\sin(2x)|\le|2yx|\]But since we need \(|x\sinh(2y)|=|y\sin(2x)|\), we see that both inequalities are actually equalities. Thus, \(x\sinh(2y)=y\sin(2x)=2xy=0\), so \(x\) or \(y=0\).

Answer 13

If you assume \(x=0\), then plug it back into your original equation, and you should get that \(y=0\) as well.

Answer 14

I hope that link will help you understand this. Sorry I couldn't find a better solution :/

Answer 15

i think i understood.did u try no b