please help me

flower3 · #1 $Old$ July 3rd, 2009, 11:44 AM

(1) $\mid (2 \overline z + 5 ) (\sqrt 2 - i )\mid$ = $\sqrt 3 \mid 2 z +5 \mid$

(2) z is either real or pure imaginary $iff {z^2}= \overline{z^2}$

Prove It · #2 $Old$ July 3rd, 2009, 12:15 PM

Quote:

Originally Posted by flower3 $View Post$

(1) $\mid (2 \overline z + 5 ) (\sqrt 2 - i )\mid$ = $\sqrt 3 \mid 2 z +5 \mid$

(2) z is either real or pure imaginary $iff {z^2}= \overline{z^2}$

Are you trying to prove these statements?

If so...

2. Suppose that $z$ is not real or pure imaginary.

Then $z = x + iy, x \neq 0, y \neq 0$ .

$z^2 = (x + iy)^2$

$= x^2 - y^2 + 2ixy$

Then $\overline{z^2} = x^2 - y^2 - 2ixy \neq z^2$ since $x \neq 0$ and $y \neq 0$ .

running-gag · #3 $Old$ July 3rd, 2009, 12:16 PM

Quote:

Originally Posted by flower3 $View Post$

(1) $\mid (2 \overline z + 5 ) (\sqrt 2 - i )\mid$ = $\sqrt 3 \mid 2 z +5 \mid$

Hi

$\mid(2 \overline z + 5 ) (\sqrt 2 - i)\mid = \mid 2 \overline z + 5 \mid \:\mid \sqrt 2 - i \mid$

$\mid \sqrt 2 - i \mid = \sqrt 3$

$\mid 2 \overline z + 5 \mid = \mid \overline {2z + 5} \mid = \mid {2z + 5} \mid$