If tan^-1(a+ib) , then x is equal to

Question

from previous year iit jee question

Answer 1

We can use the fact that e^iz=cos(z)+isin(z) and e^-iz=cos(z)-isin(z), so e^iz+e^-iz=2cos(z); similarly 2isin(z)=e^iz-e^-iz, therefore 2isin(z)/2cos(z)=itan(z)=(e^iz-e^-iz)/(e^iz+e^-iz).

Therefore, itan(z)=(e^2iz-1)/(e^2iz+1), which can also be written tan(z)=-i(e^2iz-1)/(e^2iz+1) by multiplying both sides by -i.

If tan(z)=t, then z=tan^-1(t), so if t=a+ib,

t=-i(e^2iz-1)/(e^2iz+1);

let y=e^2iz, then t=-i(y-1)/(y+1), ty+t+iy-i=0, y(t+i)+t-i=0, y=e^2iz=(i-t)/(i+t).

2iz=ln((i-t)/(i+t)) and z=(1/2i)ln((i-t)/(i+t)). Since z=tan^-1(t)=(1/2i)ln((i-t)/(i+t)).

Because t=a+ib, we can substitute for t:

tan^-1(a+ib)=(1/2i)ln((i-a-ib)/(i+a+ib)).

There are ways of expressing this using hyperbolic trigonometry.

sinh(x)=½(e^x-e^-x), cosh(x)=½(e^x+e^-x), tanh(x)=sinh(x)/cosh(x)=(e^x-e^-x)/(e^x+e^-x).

If x is replaced by ix, sinh(ix)=½(e^ix-e^-ix)=½(cos(x)+isin(x)-cos(x)+isin(x))=isin(x), isin(x)=sinh(ix).

And, similarly, cosh(ix)=½(cos(x)+isin(x)+cos(x)-isin(x))=cos(x), cos(x)=cosh(ix).

isin(x)/cos(x)=itan(x)=tanh(ix).

It follows that isin(ix)=sinh(-x)=-sinh(x), sin(ix)=isinh(x); cosh(-x)=cosh(x)=cos(ix), tan(ix)=itanh(x).