1057if dat is de case den
z2+1=-(2z+2)
solving we get z=-1±√2i
but how can we find the greatest of the two?????(complex nos cant be compared)..
\hspace{-16}$If $\mathbf{\mid z^2+1 \mid =2\mid z+1\mid }$\\\\ Then find Max. value of \mathbf{z}
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9 Answers
1057there is only 1 case for real z
z2+1=2(z+1)
solving we get z=1±√2
greatest value is 1+√2...
@man111 is there any twist to the problem?
1057
262@ketan - you have made a mistake. you cant remove the modulus just like that for complex nos.
and i think he wanted greatest value of |z|
1708Yes Rishab you are right (sorry for my mistake) actually it is |z|
and answer is |z|<=7
341But there is a tighter upper bound of 3.
Since |z^2+1| \ge |z|^2-1
we have
2|z|+2 \ge 2|z+1| \ge |z|^2-1 \Leftrightarrow |z|^2 - 2|z|-3 \le 0 \Rightarrow |z| \le 3
The minimum is \sqrt 2 -1 attained when z=1-\sqrt 2
21Lol, thats what i was doing yesterday in my head, but i realized that it wont be giving ≤ 7
so ...
1708Thanks bhatt sir , Subhodip, ketan, aditiya, Rishab
I think answer given in question paper is wrong.