Question

Let t is real number.
We have a complex number Z=(t-3)+i(t-5).
Prove that |z|^2>2 or =2 ?

Answer 1

\[\left| z \right|=\sqrt{(t-3)^2 + (t-5)^2}\]\[\left| z \right|^2 = (t-3)^2+(t-5)^2\]

Answer 2

Let's see. \(|z|=\sqrt{(t-3)^2+(t-5)^2} \implies |z|^2=(t-3)^2+(t-5)^2.\) For real numbers t, this is a parabola \(=2t^2-16t+34=2(t-4)^2+2 \ge2\)

Answer 3

ya3tek el 3afyeh :D

Answer 4

\[|z|^2 = (t - 3)^2 + (t-5)^2\]
\[f(t)= t^2 + 9 - 6t + t^2 +25 - 10t\]
\[f(t)= 2t^2 -16t + 34\]
\[f'(t) = 0\]
\[4t -16 = 0\]
\[t = 4\]
\[f(t) = (4-3)^2 + (4-5)^2 = 2\]

Answer 5

\(2(t-4)^2+2 \ge2\), because \(2(t-4)^2\ge0\).

Answer 6

Minimum Value of the function is 2 Hence f(t)>=2
\[|z|^2 \le 2\]

Answer 7

Allah ye3afeek! :D

Answer 8

sorry typo\[|z|^2 \ge 2\]