Let t is real number.
We have a complex number Z=(t-3)+i(t-5).
Prove that |z|^22 or =2 ?

Question

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

lalaly · Answer

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

anonymous · Answer

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 $=2t^2-16t+34=2(t-4)^2+2 \ge2$

lalaly · Answer

ya3tek el 3afyeh :D

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Allah ye3afeek! :D

anonymous · Answer

sorry typo$$|z|^2 \ge 2$$