Is Complex Multiplication Always Equal to Zero?

In summary: x=-2y...which gives...z_1z_2=(a-2ay)+i(ay-2a)...which gives a real part of...ax-2ay^2=0...which has infinite solutions.
  • #1
seminum
12
0
Is this statement true?

[tex]

\bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

[/tex]

Thanks.
 
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  • #2
seminum said:
Is this statement true?

[tex]

\bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

[/tex]

Thanks.

No. Can you find an example of where it won't work?

Let [itex]z_1=a+bi; z_2=x+yi;[/itex] and choose x and y such that they're not both zero, and a,b are not both zero either.
 
  • #3
Mentallic said:
No. Can you find an example of where it won't work?

Let [itex]z_1=a+bi; z_2=x+yi;[/itex] and choose x and y such that they're not both zero, and a,b are not both zero either.

What happens if the polar form of complex numbers is used instead of the rectangular form?
 
  • #4
That's what I'm getting at actually. In one textbook it's presented in polar form and it says it's true in an attempt to apply Voltage Kirchhoff's law to sum of phasors.
 
  • #5
Write ##z_1 = r_1 e^{i\theta_1}## and ##z_2 = r_2 e^{i\theta_2}##. Set ##0 = z_1 z_1##. What do you get?
 
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  • #6
George Jones said:
Write ##z_1 = r_1 e^{i\theta_1}## and ##z_2 = r_2 e^{i\theta_2}##. Set ##0 = z_1 z_1##. What do you get?

Oh man you are smart! Thanks.
 
  • #7
Sorry I forget something. The Real part of the multiplication is 0, which still does not imply one of the complex numbers is 0.
 
  • #8
If ##z = re^{i\theta} = a + ib##, then ##z = 0## if and only if ##0 = r = \sqrt{a^2 + b^2}## if and only if ##0 = a = b##.
 
  • #9
I'm confused. Are we looking at the same thing here?

seminum said:
Is this statement true?

[tex]

\bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

[/tex]

Thanks.

Let [itex]z_1=z_2=1+i[/itex], then [itex]z_1z_2 = (1+i)^2 = 1+2i-1 = 2i[/itex], hence [itex]Re(z_1z_2)=0[/itex] which disproves the statement above because [itex]z_1,z_2\neq 0[/itex].
 
  • #10
Mentallic said:
I'm confused. Are we looking at the same thing here?

We are looking at the same thing, but one of us (me) isn't able to read. No matter how many times I read it, I couldn't see the ##\bf{Re}## in

seminum said:
[tex]

\bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

[/tex]

Even
seminum said:
Sorry I forget something. The Real part of the multiplication is 0, which still does not imply one of the complex numbers is 0.

still didn't clue me in.

seminum said:
Oh man you are smart! Thanks.

Apparently not. Sorry seminum and Mentallic.

I think polar form still helps (to show that the statement is incorrect).

$$z_1 z_2 = r_1 r_2 e^{i \left( \theta_1 + \theta_2 \right)}$$

This is purely imaginary for non-zero ##z_1## and ##z_1## if ##\theta_1 + \theta_2 = \pi/2 +n\pi##. Mentallic's example has ##\theta_1 = \theta_2 = \pi/4 ##. Another simple example is ##z_1 = 1## and ##z_2 = i##.
 
  • #11
Yes I could disprove the statement using rectangular form. However based on my understanding the following text says the opposite:

Electric Circuits 9th Ed by Nilsson and Riedel
Page 321, Kirchhoff's Law in the Frequency Domain
 
  • #12
What if the line after (9.40),
But ##e^{j\omega t} \neq 0##, so

is changed to
But (9.40) is true for all ##t##, so
 
  • #13
seminum said:
Is this statement true?

[tex]

\bf{Re}\{z_1 \times z_2\} = 0 \,\, and \,\, z_2 \ne 0 \Rightarrow z_1 = 0 \,\,\, where \, z_1; z_2 \in \rm{C}

[/tex]

Thanks.

If you do the (simple) algebra in rectangular form, you'll quickly find that:

$$\bf{Re}(z_1 \times z_2) = 0 \iff \bf{Re}(z_1)\bf{Re}(z_2) = \bf{Im}(z_1)\bf{Im}(z_2)$$


and that's the only condition that needs to be satisfied. Mentallic's example is merely the special case with ##\bf{Re}(z_1) = \bf{Re}(z_2) = \bf{Im}(z_1) = \bf{Im}(z_2) = 1##.
 
  • #14
Curious3141, the text has given the right result for the wrong reason. In more detail: the text has written (for ##z## independent of ##t##) ##0 = \bf{Re} \left( z e^{j\omega t} \right\}## implies that ##z=0.##

$$0 = 2\bf{Re}\left\{ z e^{j\omega t} \right\} = z e^{j\omega t} + \left( z e^{j\omega t} \right)^* = z e^{j\omega t} + z^* e^{-j\omega t}$$

What does ##t = 0 ## give?

What does ##t = \pi/\left(2\omega\right) ## give?
 
  • #15
George Jones said:
Curious3141, the text has given the right result for the wrong reason. In more detail: the text has written (for ##z## independent of ##t##) ##0 = \bf{Re} \left( z e^{j\omega t} \right\}## implies that ##z=0.##

$$0 = 2\bf{Re}\left\{ z e^{j\omega t} \right\} = z e^{j\omega t} + \left( z e^{j\omega t} \right)^* = z e^{j\omega t} + z^* e^{-j\omega t}$$

What does ##t = 0 ## give?

What does ##t = \pi/\left(2\omega\right) ## give?

Hi George, I don't have the text, but going by what you've written, if we let ##z = re^{j\theta}##, then the condition implies that ##\bf{Re}(re^{j(\omega t + \theta)}) = 0 \implies r\cos({\omega t + \theta}) = 0 \implies r = 0 \space \rm{or} \space \omega t + \theta = \frac{\pi}{2} + 2k\pi##.

Since the latter is excluded by independence between ##z \space \rm{and}\space t##, the only possibility is ##r=0 \implies z = 0##.
 
  • #16
Yes.

More explicitly, ##t=0## gives ##0=\bf{Re} \left\{ z\right\}##, and ##t = \pi/\left(2\omega\right)## gives ##0=\bf{Im} \left\{ z\right\}##.
 
  • #17
I was just looking around on pf and I came across this thinking I would understand it. I got lost on the first post :(
 
  • #18
Gotcha! Thanks.
 
  • #19
Trevo said:
I was just looking around on pf and I came across this thinking I would understand it. I got lost on the first post :(

If you have two complex numbers,

[tex]z_1=a+ib[/tex]

[tex]z_2=x+iy[/tex]

and multiply them, of course you get

[tex]z_1z_2=(a+ib)(x+iy)=ax-by+i(ay+bx)[/tex]

and [itex]Re(z_1z_2) = ax-by[/itex] is the real part of that result (just as [itex]Re(z_1)=a, Re(z_2)=x[/itex]).

So the question is asking, if [itex]z_2\neq 0[/itex] which means that x and y aren't both zero (one of them can be zero) then are we forced to choose [itex]z_1=0[/itex] so that we can end up with [itex]Re(z_1z_2)=0[/itex]?

The answer is no because as we can see, [itex]Re(z_1z_2)=ax-by=0[/itex] and for simplicity, if we choose x=y where x and y aren't zero, then we get [itex]ax-by=ax-bx=x(a-b)=0[/itex] which means we end up with a-b=0, a=b. So we don't need to choose [itex]z_1=0[/itex] to satisfy the result, we can choose [itex]z_1=a+ia[/itex] for any a that's non-zero. Also, if we didn't restrict x=y then we'd find other combinations that satisfy the result as well.

Hence my example earlier with choosing [itex]z_1=z_2=1+i[/itex].
 

1. What is complex multiplication?

Complex multiplication is a mathematical operation that involves multiplying two complex numbers, which are numbers that have both a real and an imaginary component. It is similar to multiplying two real numbers, but with the added complexity of dealing with imaginary numbers.

2. How do you perform complex multiplication?

To perform complex multiplication, you first multiply the real components of the two numbers together, and then the imaginary components. Finally, you combine the results to get the final complex product. This can be represented using the formula (a + bi) * (c + di) = (ac - bd) + (ad + bc)i, where a and c are the real components, and b and d are the imaginary components.

3. What are the properties of complex multiplication?

Some key properties of complex multiplication include the commutative property, where changing the order of the numbers being multiplied does not change the result, and the distributive property, where multiplying a complex number by a sum is the same as multiplying the number by each term in the sum and then adding the results.

4. Can you multiply a complex number by a real number?

Yes, you can multiply a complex number by a real number. The product will be a complex number with the same imaginary component as the original number, but with the real component multiplied by the real number.

5. What are some practical applications of complex multiplication?

Complex multiplication has many practical applications in fields such as engineering, physics, and economics. It is used to model and analyze systems with both real and imaginary components, such as electrical circuits and mechanical systems. It is also used in signal processing and image processing, and plays a crucial role in understanding and solving differential equations.

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