### Author Topic: To show z1z2 = 0 implies z1 = 0 or z2 = 0  (Read 4501 times)

#### Nikita Dua

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##### To show z1z2 = 0 implies z1 = 0 or z2 = 0
« on: September 24, 2018, 03:27:11 PM »
Will a prove which uses rectangular coordinates definition of z1 and z2 would be fine?

Using z1 = x1 +iy1 and z2 = x2 + iy2 and using some algebra to prove z1z2 = 0 implies z1 = 0 or z2 = 0

Barely comprehensible. Next such post will be deleted. V.I.
« Last Edit: September 24, 2018, 06:11:55 PM by Victor Ivrii »

#### Vedant Shah

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##### Re: To show z1z2 = 0 implies z1 = 0 or z2 = 0
« Reply #1 on: September 24, 2018, 07:10:07 PM »
Given:
$z_1 z_2 = 0$
$\Rightarrow |z_1 z_2| = |0|$
$\Rightarrow|z_1||z_2| = 0$
$\Rightarrow|z_1| = 0 \vee |z_2| = 0$
We know (Modulus is non-negative; Positive for non-zero numbers):
$|z| = 0 \iff z=0$
Thus:
$z_1 z_2 = 0 \iff |z_1| = 0 \vee |z_2| = 0 \iff z_1 = 0 \vee z_2 = 0$

Another method would be to use polar coordinates:
$z_1 z_2 = r_1 e^{i \theta _1} r_2 e^{i \theta _2}$
$0 =r_1 r_2 e^{i (\theta _1 + \theta _2)}$
Since $e^{z} \gt 0$ $\forall z$:
$0 = r_1 r_2 \iff r_1 = 0 \vee r_2 = 0 \iff z_1 = 0 \vee z_2 = 0$

I am sure using rectangular would also work, but it if probably more work and algebra than necessary.
« Last Edit: September 24, 2018, 07:29:09 PM by Vedant Shah »

#### Victor Ivrii

Alternatively: $z\ne 0\implies z^{-1} =\frac{\bar{z}}{|z|^2}$. Then  $(z_1\ne 0)\wedge (z_1z_2=0)\implies z_2=z_1^{-1}z_1z_2= 0$.