3substitute z1/z2 and z2/z1 in the rhs.i think this will solve
If z1 and z2 be two distict complex numbers such that z_1+z_2=\frac{z_1}{\left|z_2 \right|^{2}}+\frac{z_2}{\left|z_1 \right|^{2}}
then prove that 1+\left|z_1 \right|\left|z _2 \right|=\frac{z_1}{z_2}+\frac{z_2}{z_1}
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8 Answers
62z_1(1-1/|z_2^2|)=-z_2(1-1/|z_1^2|)
Thus Z1 adn Z2 are parallel (or antiparallel)
I hope this helps!
39sir i have a doubt in what ever we have to prove
the LHS is a real quantity and RHS is not.
so how both of them can be equated???
341how do you know that RHS is not real? esp since nishant sir has indicated that z1/z2 is real
62@bhargav.. iin such cases the only solution is when both sides are zero..
(not to say that that is the case here)
11I've been able to prov e that z1 & z2can't be anti-parallel.....they have to be parallel...
11Continuing from where I left last night, z1=keiθ
z2=leiθ
Putting them in the given eqn we have 1k+1l=k+l, from which we have kl=1....
Let k<1 and l>1.....
but if that's the case then R.T.P requires both k and l to be equal....
But that's not possible as both z1 & z2 are distinct.......So eureka - just check ur qsn once...[99]
