# Very difficult complex no question

1. Nov 27, 2005

### shaiqbashir

very difficult complex no question plzzzz help

Hi Guys!
well! im 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 im 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 , im unable to transform it into

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

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

PLZPZLPZLPZLZ

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

Thanks and Good Bye

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2. Nov 27, 2005

### Tide

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

3. Nov 28, 2005

### shaiqbashir

Im still unable to solve this
im so near to the required answer but couldnt solve. Can you tell me if there is any special thing to keep in mind while solving this question.

4. Nov 28, 2005

### shaiqbashir

see if im correct here

Hi guys!

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 im unable to solve from this point.

what should i do?

plz help me

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

Last edited: Nov 28, 2005
5. Nov 28, 2005

### finchie_88

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

6. Nov 28, 2005

### Tide

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

7. Aug 7, 2008

### soundvslight

Re: very difficult complex no question plzzzz help

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

8. Aug 7, 2008

### soundvslight

Re: very difficult complex no question plzzzz help

see it nd tell if u need soln