Very difficult complex no question

[shaiqbashir]

very difficult complex no question pleasezzz help

Hi Guys!
well! I am facing problems in the following question:
IF
$$\tan{x+iy} = \alpha + i\beta$$
prove that
$$\alpha^2 + \beta^2 = \frac{\cosh{(h^2)(y)}-(\cos{x}^2)}{\cosh{(h^2)(y)}-(\sin{x}^2)}$$

what I am not getting here is this that you have to prove that

$$\alpha^2 + \beta^2$$

how to transform this question to give $$\alpha^2 + \beta^2$$

now i can make it like

$$\alpha^2 - \beta^2$$

but because of that i , I am unable to transform it into

$$\alpha^2 + \beta^2$$

please help me. I have just one day remaining in my exams.

PLZPZLPZLPZLZ

Yeah! if you are unable to read the above code I am attaching an attachment as well.

Thanks and Good Bye

 

[Tide]

HINT: $(\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2$

 

[shaiqbashir]

Im still unable to solve this
im so near to the required answer but couldn't solve. Can you tell me if there is any special thing to keep in mind while solving this question.

Thanks in advance

 

[shaiqbashir]

see if I am correct here

Hi guys!

thanks for your help.

Well! i want to show you how much i have done this:

i have written this

$$\tan{(x+iy)}=\frac{\sin{(x+iy)}}{\cos{(x+iy)}}$$

so when i solve this thing

im getting an answer like this

$$\tan{(x+iy)}=\frac{\sin{(x+iy)}}{\cos{(x+iy)}}= \frac{\sin{2x}+i\sinh{2y}}{\cos{2x}+\cosh{2y}}=\alpha + i\beta$$


now according to Mr. Tide I should write

$(\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2$

it means that

$$\alpha - i\beta= \frac{\sin{2x}-i\sinh{2y}}{\cos{2x}+\cosh{2y}}$$

and than i can do it like

$(\alpha + i \beta)(\alpha - i \beta) = \alpha^2 + \beta^2$

but I am unable to solve from this point.

what should i do?

please help me

i have just one day left now and on wednesday i have my exams.

Thanks in advance

 

[finchie_88]

Have you tried writing the same thing in terms of e's, that might work.
example:
$$cosh(x) = \frac{1}{2}(e^x + e^(-x))$$

 

[Tide]

HINT: Multipley $\tan (x+iy) \tan (x-iy)$ and use $\sin z = (e^{iz} - e^{-iz})/(2i)$ and similarly for the cosine.

 

[soundvslight]

hi shaikh
i think d q s of coll level
it includes circular and hyperbolic fncns

 

[soundvslight]

see it nd tell if u need soln