## mathslover 2 years ago Let z be a non-real complex number lying on the circle |z| = 1. Then prove that : z = $$\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}$$

1. mathslover

wait

2. mathslover

$\large{z = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}}$

3. modphysnoob

non-real complex? do you mean just imaginary?

4. ParthKohli

In which grade do you learn this?

5. mathslover

6. mathslover

I was able to prove by taking RHS and simplifying and simplifying it into such a form which was exactly equivalent to the LHS i.e. 'z' But , the book says this is not a method of proving infact it is a method of "verifying" ... :(

7. mathslover

:P Well lemme post what I think,,,, it will take time...

8. ParthKohli

(nothing to do here)

9. JamesJ

First thing to do is rationalize the denominator. Multiply top and bottom by 1 + i . tan(Arg(z)/2)

10. AravindG

you can also do this by taking modulus of given equation

11. AravindG

|dw:1360507411970:dw|

12. AravindG

you will get magnitude as 1

13. mathslover

@JamesJ sir but that will be verifying --- my book as well as the teacher says

14. mathslover

@AravindG yes I tried that and got : $\large{\frac{z}{\cos \theta} = 1 + i \tan \theta }$

15. AravindG

how u gt that ??

16. AravindG

I have 1 more method in stock ..bt I am keeping it as a novelty for end of discussion Let me see your working here

17. mathslover

@AravindG : Say : z = $$\cos \theta + i \sin \theta$$ Divide both sides by cos theta $$\large{\frac{z}{\cos \theta} = \frac{\cos \theta + i\sin \theta}{\cos \theta}}$$

18. mathslover

^ This will give us : $\large{\frac{z}{\cos\theta} = 1 + i\tan\theta}$

19. mathslover

which I can also write as : $\large{\frac{z}{\cos\theta} = 1 + i \tan (arg(z))}$

20. mathslover

ok so I divide both sides by 2 now... lemme do it now .. I am curious that I can do it just in the guidance of you all

21. AravindG

of course we are watching..do ur best

22. Spacelimbus

$\Large \frac{1}{\cos\phi}z*=1-i\tan\phi$

23. Spacelimbus

Oh sorry, I was slow in my typing, you got that far already.

24. AravindG

complex number is one of my fav topics ...do tag me next time u hav a qn on it

25. mathslover

Ok so I got : $\large{\frac{ z + \cos \theta}{z \cos \theta} = 1 + i \tan \frac{\theta}{2}}$

26. AravindG

why did you do that??

27. AravindG

you are looking for z , just equate for z rest will come up :)

28. mathslover

To get the numerator (see the RHS of required to prove equation) ... now I will go for denominator

29. JamesJ

What I am suggesting is this. Let the left hand side be called $$w$$. Then $w = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})} . \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 + i \tan ( \frac{arg(z)}{2})}$ $= \frac{ (1 + i \tan ( \frac{ arg(z)}{2} ))^2}{1 + \tan^2(\arg(z)/2)}$ $= \frac{1 - \tan^2(\arg(z)/2) + 2i \tan(\arg(z)/2)}{\sec^2(\arg(z)/2)}$ Then the tan of the argument of $$w$$ is $\tan(\arg(w)) = \frac{2 \tan(\arg(z)/2)}{1 - \tan^2(\arg(z)/2)} = \tan(\arg(z))$ and hence $$w$$ and $$z$$ have the same same arguments Now you just need to show that $$|w| = |z|$$ and then it will follow that $$w = z$$.

30. mathslover

OK mistake there it is : $\large{\frac{z + \cos\theta}{2\cos\theta} = 1 + i \tan \frac{\theta}{2}}$

31. AravindG

still y did you go for this ? you could have got it from 1st eqn itself using $$\tan 2 \theta$$ formula

32. mathslover

and then I will subtract $$i\tan\frac{\theta}{2}$$ both sides so that I get : $\large{\frac{z+ \cos\theta - 2i\tan\theta}{2\cos\theta} = 1 -i\tan\frac{\theta}{2}}$

33. AravindG

ok nvm u can continue

34. mathslover

Oh yes @AravindG let me try that ... great thanks but dont' leave now , I am yet to finish :)

35. mathslover

Oh! Sorry everyone got to go now.. but I promise I will be back and write my solution what I get and @AravindG thanks for that trick^^^ and @JamesJ sir I will get back to you too what I not understand from your soln ... Sorry friends but it is urgent Thanks everyone

36. AravindG

ok all the best !

37. AravindG

i will post a better method if I get one

38. AravindG

ok got it :)

39. AravindG

Let the complex number be $\bf \large z=\cos \theta +i \sin \theta$ $\bf \large z=\cos^2 \theta-\sin^2 \theta +i(2\sin \dfrac{\theta}{2}\cos \dfrac{\theta}{2})$ $\bf \large z=(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})^2$ $\bf \large =(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})$ =$\bf \large =\dfrac{\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}-i \sin \dfrac{\theta}{2}}$ $\bf \large =\dfrac{1+ i \tan \dfrac{\theta}{2}}{1- i \tan \dfrac{\theta}{2}}$ $\bf \large z=\dfrac{1+i \tan \dfrac{\arg(z)}{2}}{1+i \tan \dfrac{\arg(z)}{2}}$

40. AravindG

oh in second step its $\large \cos ^2\dfrac{\theta}{2}-\sin^2 \dfrac{\theta}{2}$

41. JamesJ

Yes, that's good.

42. AravindG

Hence proved :)