Question

I have a question regarding complex numbers:

u=2+i
u*=2-i

By considering the argument of (u/u*), or otherwise, prove that tan^-1(4/3)=2tan^-1(1/2)

Answer 1

okay let's solve it

what is value of (u/u*)? can you do it?

Answer 2

Sorry for the late reply.

Yes, I know the answer is (4+4i-1)/5 , I can't solve the rest.

Answer 3

let's do it...
your answer is correct...

u/u*=\[(4+4i-1)/5 =(3+4i)/5\] \[=3/5+4/5i\]

now what is argument of above complex no.?

by comparing to standard form of complex no. z=x+yi

we get x=3/5 and y=4/5

argument theta of (u/u*) is

\[\theta=\tan^{-1} (y/x)\]

\[\theta=\tan^{-1} (4/3)\] .................(1)

now you can covert given complex nos.u and u* in polar form
of (r,theta) [r=modulus of complex no. and theta is argument of complex no.]

you get,

\[u=(\sqrt{3},\theta _{1})\] and

\[u*=(\sqrt{3},\theta _{2})\]

where \[\theta _{1}=\tan ^{-1}(1/2)\]

also \[\theta_{2}=\tan^{-1} (-1/2)=-\tan^{-1}(1/2) \]
(as we know tan^-1(-x)=-tan^-1(x) )

now what we have to do.. we have to find (u/u*)

now let's to this division in polar form

now here is key

when we do division in polar form the resultant complex no. has
magnitude =magnitude of complex no . in numerator / magnitude of complex no. in denominator

and resultant argument is=argument of complex no. in numerator - argument of complex no. in denominator
that is

if \[a=(r _{1},\theta _{1}) \] and
\[b=(r _{2},\theta _{2})\]

magnitude of complex no. (a/b)=(r1/r2)

and argument is = (theta1 -theta2)

now here we have a=u and b=u*
and so

magnitude of complex no (u/u*) is

\[\sqrt{3}/\sqrt{3}=1\]

and resultant argument of (u/u*) is

\[\theta _{1}-\theta _{2}=\tan^{-1} (1/2)- [-\tan^{-1} (1/2)=\tan^{-1} (1/2)+\tan^{-1}(1/2)=2\tan^{-1} (1/2)\]..........(2)


so from results (1) and (2) it is proved that

tan^-1(4/3)=2tan^-1(1/2) !!!

i hope this helps :)

Answer 4

Thanks! The answer helped a lot!

Answer 5

you are most welcome...! :)