Product of two complex numbers = 0

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  • #1
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

The Attempt at a Solution


I have absolutely no clue.
 

Answers and Replies

  • #2
clamtrox
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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 = ?
 
  • #3
Hurkyl
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Hint: try working in exponential polar form
Have you tried that yet? What does the question look like in exponential polar form?
 
  • #4
vesu
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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?
 
  • #5
Whovian
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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?

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.
 
  • #6
cepheid
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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?


Correct, except what does them being negative have to do with anything?

Edit: what I mean is, correct answer, totally wrong reasoning.
 
  • #7
vesu
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Correct, except what does them being negative have to do with anything?

Edit: what I mean is, correct answer, totally wrong reasoning.
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

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.
[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] ?
 
  • #8
clamtrox
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[itex]e^{i(\theta + \phi)} > 0[/itex]
It certainly can be smaller than 0, but can it be exactly 0?
 
  • #9
vesu
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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] ?

It certainly can be smaller than 0, but can it be exactly 0?
Oh yeah, so it should be [itex]e^{i(\theta + \phi)} \not= 0[/itex] then?

I think it's all correct other than that, right?
 
  • #10
Whovian
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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

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?
 
  • #11
Ray Vickson
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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

The Attempt at a Solution


I have absolutely no clue.

You can also do it directly: if z = x + i*y and w = u + i*v with x,y,u,v real, then
zw = (xu - vy) + i*(xv + yu), so we need xu = vy and xv = -uy. You can consider several (easy) cases such as u =/= 0 vs, u = 0, and so conclude that in the first case you *must have* x = y = 0, while in the second case you have either v = 0 (so you are done) or else you have x = y = 0.

RGV
 
Last edited:
  • #12
vesu
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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?
I'm a little confused, could you explain how this is incorrect?

[itex]zw = 0[/itex]
[itex]zw = re^{i\theta} \times qe^{i\phi}[/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]
[itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]

I guess you could also do it this way to eliminate [itex]i[/itex] altogether so that you're only working with reals? This doesn't involve exponential polar form though so I'm not sure if this is how we're expected to solve it.

[itex]z = a + ib[/itex]
[itex]w = c + id[/itex]
[itex]|zw| = |z||w| = 0[/itex]
[itex]\sqrt{a^2 + b^2} \times \sqrt{c^2 + d^2} = 0[/itex]
[itex]\therefore a[/itex] and [itex]b = 0[/itex]
or [itex]c[/itex] and [itex]d = 0[/itex]
[itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]
 
Last edited:
  • #13
Dickfore
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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]
 
  • #14
vesu
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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]
This has confused me even more. :S

also I cringe every time I re-read this:
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]
I was very very tired. :frown:
 
  • #15
Whovian
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[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]

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.
 
  • #16
HallsofIvy
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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
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.

[itex]rqe^{i(\theta + \phi)}[/itex] ?

and then since [itex]e^{i(\theta + \phi)} > 0[/itex]
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.

either [itex]r=0[/itex] or [itex]q=0[/itex], therefore either [itex]z = 0[/itex] or [itex]w = 0[/itex] ?
It is, however, true that [itex]e^{ix}[/itex], for x real, is never 0.
 
  • #17
vesu
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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. :confused:

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.
 
  • #18
Whovian
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Try taking the absolute value of both sides.
 
  • #19
Ray Vickson
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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. :confused:

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.

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
 
  • #20
vela
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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.
 
  • #21
Whovian
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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.

Well, the step I was quoting assumes that, for some real numbers r, q, and ##\psi##,

$$r\cdot q\cdot e^{\psi\cdot i}=0$$

and says that

$$r = 0\lor q = 0\lor e^{\psi\cdot i}=0$$

As ##e^{\psi\cdot i}## isn't necessarily real, this step's invalid. There are a few ways, however, to proceed from here. My favourite is taking the absolute value of both sides, but one could also write ##e^{\psi\cdot i}## as ##a+b\cdot i## for real numbers a and b and work from there.
 
  • #22
vela
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But vesu didn't do that. Vesu noted the exponential isn't 0, which means it could be divided it out, leaving rq=0.
 
  • #23
cepheid
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I'm confused too. In the OP's proof, the claim that the complex exponential factor cannot be 0 seems sufficient. We're you looking for the OP to say explicitly that:

|exp[i(theta + phi)]|

= |1*exp[i(theta + phi)]|

= 1

And therefore the complex exponential part can never be 0, because it must lie on the unit circle in the complex plane?
 
  • #24
clamtrox
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But vesu didn't do that. Vesu noted the exponential isn't 0, which means it could be divided it out, leaving rq=0.

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.
 
  • #25
vesu
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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
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!

Is this proof okay if the other one isn't? (even though this doesn't involve exponential polar form, that was only a hint, not a requirement)

let [itex]z = a + ib[/itex] and [itex]w = c + id[/itex]

[itex]zw = 0[/itex]
[itex]|zw| = 0[/itex]
[itex]|z||w| = 0[/itex]
[itex]\sqrt{a^2 + b^2} \times \sqrt{c^2 + d^2} = 0[/itex]
[itex]\therefore a[/itex] and [itex]b = 0[/itex]
or [itex]c[/itex] and [itex]d = 0[/itex]
[itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]

Thoroughly confused at this point. :confused:
 
  • #26
clamtrox
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[itex]|zw| = 0[/itex]
[itex]|z||w| = 0[/itex]

This step is what you want to prove.
 
  • #27
vesu
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This 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
 
  • #28
clamtrox
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Is it? I was just using [itex]|zw|=|z||w|[/itex], I don't think that's what we're trying to prove. :S

It is. That's what the whole brouhaha has been in this thread :) Think about it for a while!
 
  • #29
vela
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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.
If you know, for example, that ℂ is a field, then yeah, this works. Dickfore suggested this method earlier in the thread.
 
  • #30
vela
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I'm confused too. In the OP's proof, the claim that the complex exponential factor cannot be 0 seems sufficient.
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≠0, then ze=0 implies z=0 because e 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.
 
  • #31
vesu
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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≠0, then ze=0 implies z=0 because e 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.
so

[itex]zw = 0[/itex]
[itex]re^{i\theta} \times qe^{i\phi} = 0[/itex]
[itex]rqe^{i(\theta + \phi)} = 0[/itex]
[itex]e^{i(\theta + \phi)} \not= 0[/itex]
[itex]\frac{rqe^{i(\theta + \phi)}}{e^{i(\theta + \phi)}} = \frac{0}{e^{i(\theta + \phi)}}[/itex]
[itex]rq = 0[/itex]
[itex]\therefore r=0[/itex] or [itex]q=0[/itex]
[itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]

Um, I can't find anything at all in our course material about taking the modulus of a complex exponential. According to WolframAlpha [itex]|e^{i\theta}| = 1[/itex] assuming [itex]\theta[/itex] is real? So, uh...

[itex]zw = 0[/itex]
[itex]re^{i\theta} \times qe^{i\phi} = 0[/itex]
[itex]rqe^{i(\theta + \phi)} = 0[/itex]
[itex]|rqe^{i(\theta + \phi)}| = |0|[/itex]
[itex]rq = 0[/itex]
[itex]\therefore r=0[/itex] or [itex]q=0[/itex]
[itex]\therefore z = 0[/itex] or [itex]w = 0[/itex]

Both of these seem okay I think?
 
  • #32
clamtrox
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According to WolframAlpha [itex]|e^{i\theta}| = 1[/itex] assuming [itex]\theta[/itex] is real?

True, but you can see fairly easily again that this claim is equivalent with claiming that |zw|= |z| |w|. Maybe you can somehow convince yourself (without resorting to wolframalpha) that this is true?
 
  • #33
cepheid
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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≠0, then ze=0 implies z=0 because e 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.

What is a "zero divisor?"
 
  • #35
cepheid
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