Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

wait

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

non-real complex? do you mean just imaginary?

In which grade do you learn this?

@ParthKohli in 11th grade

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

(nothing to do here)

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

you can also do this by taking modulus of given equation

|dw:1360507411970:dw|

you will get magnitude as 1

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

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

how u gt that ??

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

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

of course we are watching..do ur best

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

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

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

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

why did you do that??

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

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

ok nvm u can continue

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

ok all the best !

i will post a better method if I get one

ok got it :)

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

Yes, that's good.

Hence proved :)