- #1
vesu
- 19
- 0
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 = reiθ
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?vesu said:Hint: try working in exponential polar form
clamtrox said: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 = ?
vesu said: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?
vesu said: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? :Scepheid said:Correct, 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] ?Whovian said: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?vesu said:[itex]e^{i(\theta + \phi)} > 0[/itex]
vesu said: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?clamtrox said:It certainly can be smaller than 0, but can it be exactly 0?
vesu said: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
vesu said: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 = reiθ
The Attempt at a Solution
I have absolutely no clue.
I'm a little confused, could you explain how this is incorrect?Whovian said: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. :SDickfore said:If [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.vesu said: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]
vesu said:[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.vesu said: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.Whovian said: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.
vesu said: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.Whovian said: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.
vela said: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.
vela said: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!Ray Vickson said: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
vesu said:[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. :Sclamtrox said:This step is what you want to prove.
vesu said: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.clamtrox said: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 eiθ≠0, then zeiθ=0 implies z=0 because eiθ 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.cepheid said:I'm confused too. In the OP's proof, the claim that the complex exponential factor cannot be 0 seems sufficient.
sovela said: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 eiθ≠0, then zeiθ=0 implies z=0 because eiθ 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.
vesu said:According to WolframAlpha [itex]|e^{i\theta}| = 1[/itex] assuming [itex]\theta[/itex] is real?
vela said: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 eiθ≠0, then zeiθ=0 implies z=0 because eiθ 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.
clamtrox said:
When the product of two complex numbers equals 0, it means that the two numbers are orthogonal or perpendicular to each other on the complex plane. This means that the two numbers are at a 90 degree angle from each other and their product is 0.
No, the product of two non-zero complex numbers cannot equal 0. This is because the product of two non-zero numbers on the complex plane will always have a non-zero magnitude and a non-zero angle, resulting in a non-zero product.
To solve for the two complex numbers when their product equals 0, you can set up an equation where the product of the two numbers is equal to 0. Then, you can solve for each number individually by setting one of them equal to 0 and solving for the other number.
Yes, the product of a real number and an imaginary number can equal 0. This can occur when the real number is 0, resulting in a product of 0 regardless of the value of the imaginary number.
Yes, the concept of orthogonality in complex numbers is used in fields such as engineering and physics. For example, in electrical engineering, the real and imaginary components of a complex impedance are orthogonal and their product is 0, allowing for easier analysis of circuits.