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vesu
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Homework Statement
Let z and w be two complex numbers such that zw = 0. Show either z = 0 or w = 0.
Hint: try working in exponential polar form
Homework Equations
z = re^{iθ}
The Attempt at a Solution
I have absolutely no clue.
Have you tried that yet? What does the question look like in exponential polar form?Hint: try working in exponential polar form
What happens if you write two complex numbers in polar form, then multiply them together and require the result is equal to zero? So [itex] z=re^{i\theta}, w = qe^{i\phi} [/itex],zw = ?
so [itex]zw = re^{i\theta} \times qe^{i\phi}[/itex]
and since neither [itex]re^{i\theta}[/itex] or [itex]qe^{i\phi}[/itex] can be negative, either [itex]z = 0[/itex] or [itex]w = 0[/itex]. Is this correct?
so [itex]zw = re^{i\theta} \times qe^{i\phi}[/itex]
and since neither [itex]re^{i\theta}[/itex] or [itex]qe^{i\phi}[/itex] can be negative, either [itex]z = 0[/itex] or [itex]w = 0[/itex]. Is this correct?
Right, whoops. So either z = 0 or w = 0 because of the zero-product property? i.e. "if ab = 0, then either a = 0 or b = 0"? Why is working in exponential polar form at all necessary or even helpful then? :SCorrect, except what does them being negative have to do with anything?
Edit: what I mean is, correct answer, totally wrong reasoning.
[itex]rqe^{i(\theta + \phi)}[/itex] ?Not in the slightest. Does negative mean anything when it comes to complex numbers? Also, when imaginary exponents are involved, [itex]e^{a+b\cdot i}[/itex] can be anything. Basic identities about exponents, one of which should be screaming at you, still apply.
It certainly can be smaller than 0, but can it be exactly 0?[itex]e^{i(\theta + \phi)} > 0[/itex]
Right, whoops. So either z = 0 or w = 0 because of the zero-product property? i.e. "if ab = 0, then either a = 0 or b = 0"? Why is working in exponential polar form at all necessary or even helpful then? :S
[itex]rqe^{i(\theta + \phi)}[/itex] ?
and then since [itex]e^{i(\theta + \phi)} > 0[/itex] either [itex]r=0[/itex] or [itex]q=0[/itex], therefore either [itex]z = 0[/itex] or [itex]w = 0[/itex] ?
Oh yeah, so it should be [itex]e^{i(\theta + \phi)} \not= 0[/itex] then?It certainly can be smaller than 0, but can it be exactly 0?
Right, whoops. So either z = 0 or w = 0 because of the zero-product property? i.e. "if ab = 0, then either a = 0 or b = 0"? Why is working in exponential polar form at all necessary or even helpful then? :S
Homework Statement
Let z and w be two complex numbers such that zw = 0. Show either z = 0 or w = 0.
Hint: try working in exponential polar form
Homework Equations
z = re^{iθ}
The Attempt at a Solution
I have absolutely no clue.
I'm a little confused, could you explain how this is incorrect?The point is that we already know it works for reals. Let's assume it works for reals. Now prove it works for complex numbers. Relying on what you're trying to prove won't get you anywhere. If you really do need a hint,
Ever heard of the absolute value function and how it works on complex numbers?
This has confused me even more. :SIf [itex]z = 0[/itex], you are done. Suppose [itex]z \neq 0[/itex]. Then, the multiplicative inverse of z exists. But, then:
[tex]
w = \frac{1}{z} \cdot 0 = ?
[/tex]
I was very very tired.so [itex]zw = re^{i\theta} \times qe^{i\phi}[/itex]
and since neither [itex]re^{i\theta}[/itex] or [itex]qe^{i\phi}[/itex] can be negative, either [itex]z = 0[/itex] or [itex]w = 0[/itex]
[itex]zw = rqe^{i(\theta + \phi)}[/itex]
[itex]e^{i(\theta + \phi)} \not= 0[/itex]
[itex]\therefore r=0[/itex] or [itex]q=0[/itex]
Because this is what you are being asked to prove! That is, you are asked to prove that the "zero product property" holds for complex numbers as well as real numbers.Right, whoops. So either z = 0 or w = 0 because of the zero-product property? i.e. "if ab = 0, then either a = 0 or b = 0"? Why is working in exponential polar form at all necessary or even helpful then? :S
NO! This not true. For example, [itex]e^{i\pi}= -1[/itex]. But, as you were told before, there is no "less than" or "larger than" relation on the complex numbers.[itex]rqe^{i(\theta + \phi)}[/itex] ?
and then since [itex]e^{i(\theta + \phi)} > 0[/itex]
It is, however, true that [itex]e^{ix}[/itex], for x real, is never 0.either [itex]r=0[/itex] or [itex]q=0[/itex], therefore either [itex]z = 0[/itex] or [itex]w = 0[/itex] ?
Uh... I understand what you're saying about what's wrong with my proof but I really don't know what to do even with the hint.It's this step. To justify it, we need to rely on the very identity we are trying to prove, resulting in circular reasoning. Hint: find some way to introduce an absolute value into this equation.
Uh... I understand what you're saying about what's wrong with my proof but I really don't know what to do even with the hint.
I mean I know [itex]r = |z|[/itex] and [itex]q = |w|[/itex] so [itex]rq=|zw|[/itex] but I don't know where that gets me.
It's not circular reasoning. It's safe to assume here that for real numbers x and y, that xy=0 implies that x=0 or y=0, and this is what the proof is relying on as r and q are both real numbers. The OP is being asked to show the same property holds for complex numbers.It's this step. To justify it, we need to rely on the very identity we are trying to prove, resulting in circular reasoning. Hint: find some way to introduce an absolute value into this equation.
It's not circular reasoning. It's safe to assume here that for real numbers x and y, that xy=0 implies that x=0 or y=0, and this is what the proof is relying on as r and q are both real numbers. The OP is being asked to show the same property holds for complex numbers.
But vesu didn't do that. Vesu noted the exponential isn't 0, which means it could be divided it out, leaving rq=0.
Sorry, I'm not really sure what to do with your suggestion and I'm still trying to figure out if either of the proofs I have is correct!I gave you an alternative suggestion in my previous posting. Have you looked at it? Have you tried to complete the argument I gave?
RGV
[itex]|zw| = 0[/itex]
[itex]|z||w| = 0[/itex]
Is it? I was just using [itex]|zw|=|z||w|[/itex], I don't think that's what we're trying to prove. :SThis step is what you want to prove.
Is it? I was just using [itex]|zw|=|z||w|[/itex], I don't think that's what we're trying to prove. :S
If you know, for example, that ℂ is a field, then yeah, this works. Dickfore suggested this method earlier in the thread.Would it be valid then to reason that
zw = 0
case 1: z≠0
divide by z
∴w = 0
case 2: w≠0
divide by w
∴z = 0.
case 3: w=z=0
0*0 = 0.
Actually, it isn't, which is Whovian's point. Since we're trying to prove that there are no zero divisors, so we can't assume that just because e^{iθ}≠0, then ze^{iθ}=0 implies z=0 because e^{iθ} might be a zero divisor. You should show how to get rid of the complex exponential, e.g. take the modulus or multiply by e^{-iθ}.I'm confused too. In the OP's proof, the claim that the complex exponential factor cannot be 0 seems sufficient.
soActually, it isn't, which is Whovian's point. Since we're trying to prove that there are no zero divisors, so we can't assume that just because e^{iθ}≠0, then ze^{iθ}=0 implies z=0 because e^{iθ} might be a zero divisor. You should show how to get rid of the complex exponential, e.g. take the modulus or multiply by e^{-iθ}.
Now I agree that the OP was likely using circular reasoning while the rest of us glossed over it, thinking "Obviously you can get rid of the exponential." It was a good catch by Whovian.
According to WolframAlpha [itex]|e^{i\theta}| = 1[/itex] assuming [itex]\theta[/itex] is real?
Actually, it isn't, which is Whovian's point. Since we're trying to prove that there are no zero divisors, so we can't assume that just because e^{iθ}≠0, then ze^{iθ}=0 implies z=0 because e^{iθ} might be a zero divisor. You should show how to get rid of the complex exponential, e.g. take the modulus or multiply by e^{-iθ}.
Now I agree that the OP was likely using circular reasoning while the rest of us glossed over it, thinking "Obviously you can get rid of the exponential." It was a good catch by Whovian.