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In school, I was taught that complex numbers could be represented using vectors. My teacher used this technique on sample problems. It had never failed me and I've been able to use it to solve many problems involving complex numbers. However, today I realized I didn't understand what I was doing.

Consider a complex number $z = 3 + 4i.$ Let $M$ be the point representing $z$ on the complex plane $Oxy.$ I was taught that $|z| = |\vec{OM}| = \sqrt{3^2 + 4^2} = 5.$ This is certainly true for $z,$ and I could visualize why this is also true in terms of vector $\vec{OM}$.

Likewise, $|z^2| = |\vec{OM}\cdot\vec{OM}| = OM^2 = 3^2 + 4^2 = 25.$

Let $N$ represent the complex number $w.$ We also did operations like:

$|z - w| = |\vec{OM} - \vec{ON}| \implies |z - w|^2 = |\vec{OM} - \vec{ON}|^2 = OM^2 + ON^2 - 2\vec{OM}\cdot\vec{ON} $

This way, suppose we know the values of $|z - w|, OM, ON,$ we could find the value of $\vec{OM}\cdot\vec{ON}.$

However, today I just found that doing this could lead to weird results. Consider two complex numbers $a = 1$ and $b = i,$ represented by $A$ and $B$ on the complex plane $Oxy.$ Then:

$|a \cdot b| = |1 \cdot i| = 1$

However, $|\vec{OA}\cdot\vec{OB}| = 0$

So obviously, $|a \cdot b| \ne |\vec{OA}\cdot\vec{OB}|$

Given my inability to explain this, I clearly have not understood the true nature of what I have been doing. I hope to receive some advice on this.

My questions: What makes the vector representations work in all the first examples but not in the last example? To what extent could we use vector representation to produce an operation equivalent to one in complex number algebra?

Thank you very much!

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    $\begingroup$ I am not sure what you are asking, but it seems like you discovered that in the vector representation of complex numbers, multiplication does not correspond to the dot product of the relevant vectors. This is certainly true...there is not much more to say about it? $\endgroup$
    – Carlyle
    Commented May 24 at 10:33
  • $\begingroup$ @Carlyle Hi, thank you! I'm mainly interested in knowing why the example I showed falls apart instead of just accepting that it doesn't work in that case but will work in some others. Also, I did provide an example where multiplication does correspond to the dot product of the related vectors. ie. $|z^2| = |z \cdot z| = |\vec{OM}\cdot\vec{OM}|.$ Apparently, this no longer seems to hold in my last example. $\endgroup$ Commented May 24 at 10:39
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    $\begingroup$ The dot product $(a+ib)\cdot (c+id)=ac+bd$. Complex multiplication yields $(ac-bd)+(bc+ad)i$. They are two different operations. Why would they give the same result? $\endgroup$
    – John Douma
    Commented May 24 at 11:06
  • $\begingroup$ @JohnDouma I believe I did not do the good job clarifying my confusion. I understand that they are different operations, one for complex number algebra and another for vectors. However, as explained in the post, we were taught that we could turn an operation in complex number algebra into an equivalent operation in vectors. It has worked, but it doesn't work in the last example I showed. So my question was what's the difference between the last example and the other examples? What causes it to fail? $\endgroup$ Commented May 24 at 11:14
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    $\begingroup$ The thing you are missing is that the dot product on vectors is not equivalent to multiplication on complex numbers. You must consider the context of what you are being taught. The equivalent operation for the dot product on vectors is the dot product on complex numbers where $1=(1,0)$ and $i=(0,1)$ are the basis vectors for the complex numbers as a vector space. $\endgroup$
    – John Douma
    Commented May 24 at 12:29

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The heart of your issue is this: the dot product of two complex numbers and the usual product of complex numbers is not the same product. Full stop. There is no "in some cases it's the same and in others it isn't"*. If you are ever exchanging a dot product for a normal product or vice versa, you are most likely making a mistake.

The only algebraic operation on the complex numbers that corresponds to a geometric operation on vectors is the sum of complex numbers, which corresponds to the vector sum.

* Technically speaking, it is the same for the boring case where all numbers involved are purely real. But there's no need to bring vector geometry into it then, because then everything happens on a 1d line anyway.

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  • $\begingroup$ Thanks! I think I'm close to getting it. I'd ask you the same question I posted in the other answer: Does the case where $|z^2| = |z \cdot z| = |\vec{OM}\cdot\vec{OM}|$ happens to be correct because we have two vectors pointing in the same direction? ie. their $cos\phi = 1$? $\endgroup$ Commented May 24 at 11:41
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    $\begingroup$ That case is an artifact of how the modulus is related to both complex multiplication and the dot product, rather than a feature of complex and dot product. It comes from the facts that $z\bar z=|z|^2$ and $\vec v\cdot\vec v=|\vec v|^2$. So $z\bar z=\overrightarrow{OM}\cdot\overrightarrow{OM}$, and this is indeed true if and only if we consider vectors/numbers pointing in the same direction, since $u\bar v=\vec u\cdot\vec v+iu\wedge v$, where $\wedge$ is a 2d equivalent of the cross product and is 0 iff $u||v$. $\endgroup$ Commented May 24 at 13:55
  • $\begingroup$ Thank you very much! $\endgroup$ Commented May 24 at 15:43
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I prefer to discuss the issue differently. Given any complex number $~z \in \Bbb{C},~$ you can uniquely identify $~x,y \in \Bbb{R},~$ such that $~z = x + iy ~: ~i = \displaystyle \sqrt{-1}.$

Note that $~z = 0 = 0 + i(0),~$ if and only if $~[ ~x = 0 ~~\text{and} ~~y = 0 ~].$

Here, I am going to assume that angles are measured in degrees, rather than radians, and that for any angle, there is a unique value $~\theta~$ that is in the half open interval of $~[0^\circ, 360^\circ),~$ such that the angle is equal to $~\theta.~$

Note that if the complex number $~z = x + iy~$ is not equal to $~0~$ then the value $~\displaystyle r = \sqrt{x^2 + y^2}~$ must be a positive number.

Further, given any $~z = x + iy \neq 0,~$ with $~r~$ defined as in the previous paragraph, you have that $~\displaystyle \left[ ~\frac{x}{r} ~\right]^2 + \left[ ~\frac{y}{r} ~\right]^2 = 1.~$

This implies that for any $~z = x + iy \neq 0,~$ you can identify a unique angle $~\theta \in [0^\circ, 360^\circ),~$ such that $~\dfrac{x}{r} = \cos(\theta),~$ and $~\dfrac{y}{r} = \sin(\theta).~$

What this implies is that given any $~z = x + iy \neq 0,~$ you can uniquely associate $~z~$ with the two component vector of $~(r,\theta),~$ such that

  • $~r \in \Bbb{R^+}~~$ and

  • $~\theta \in [0^\circ, 360^\circ).$

So, for any $~z = x + iy \neq 0,~$ you can construe the associated components of $~r~$ as the magnitude of $~z,~$ and $~\theta~$ as the direction of $~z.~$

I realize that this is somewhat off topic to the specific syntax used in the posted question. However, I have found that this representation of $~z \neq 0,~$ by the vector $~(r,\theta)~$ tends to readily expand the math student's intuition.

This intuitive translation of the syntax used in the posted question into the $~(r,\theta)~$ syntax may well make it easier for the math student to intuitively understand the syntax used in the posted question.

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  • $\begingroup$ Though I understand this does not answer the question, I really appreciate the discussion. Thank you! $\endgroup$ Commented May 24 at 12:12
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    $\begingroup$ @ten_to_tenth You may wish to consider printing out this entire webpage, including my answer, and showing it to your math teacher. This will allow the math teacher to consider altering the way that they present the topic of vectors to represent complex numbers not equal to $~0.~$ There is clearly value in the syntax used in the posted question. However, the intuitive translation (back and forth) between the $~(r,\theta)~$ syntax and the syntax used in the posted question also has value. $\endgroup$ Commented May 24 at 12:19
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Your Inconsistency

When you calculate the magnitude

$$ \left|\vec{OM}\cdot\vec{OM}\right| = 3^{2} + 4^{2} $$

you multiply the real part of the vector with the real part, the imaginary part with the imaginary part, then calculate their sum.

However, when you calculate the magnitude

$$ \left|a \cdot b\right| $$

you multiply the real part of $a$ with the imaginary part of $b$, then take the absolute value of it. Which was why you got $1$. If you multiplied the real part of $a$ with the real part of $b$, and likewise for the imaginary part, you'd get the correct answer, which is $0$.

The Relation that Holds

The following holds:

$$ \Re{\left(z\cdot \bar{z}\right)} = \vec{OM}\cdot\vec{OM} = 25 $$

$$ \Re{\left(a\cdot \bar{b}\right)} = \vec{OA}\cdot\vec{OB} = 0 $$

also, the following holds

$$ \left|z \cdot z\right| = \left|z\right|\cdot\left|z\right| = \left|\vec{OM}\right|\cdot\left|\vec{OM}\right| = 25 $$

$$ \left|a \cdot b\right| = \left|a\right|\cdot\left|b\right| = \left|\vec{OA}\right|\cdot\left|\vec{OB}\right| = 1 $$

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  • $\begingroup$ Thank you! I was simply doing normal multiplication for complex numbers. When multiplying two complex numbers together, am I supposed to multiply the real parts together and the imaginary parts together? E.g. $|(1 + 2i)(1 + 3i)| = |-5 + 5i| = 5\sqrt{2}$ instead of $|1\cdot1 + 2\cdot3| = 7.$ $\endgroup$ Commented May 24 at 10:45
  • $\begingroup$ @ten_to_tenth in your example, $5\sqrt{2}$ is the magnitude of the product between two complex numbers, while $7$ is the inner product of their vector representations. So it depends on what you are trying to compute $\endgroup$
    – acat3
    Commented May 24 at 10:50
  • $\begingroup$ I just updated my questions if I weren't being clear initially. $|a \cdot b|$ is the modulus of the complex number that is $a \cdot b,$ so I believe it could not be the inner product you were referring to? What confuses me is why sometimes we could use vectors to produce an operation equivalent to the original operation in complex number algebra, but could not in some other situations. $\endgroup$ Commented May 24 at 11:00
  • $\begingroup$ E.g. $|z^2| = |z \cdot z| = |\vec{OM}\cdot\vec{OM}|,$ but $|a \cdot b| \ne |\vec{OA}\cdot\vec{OB}|$ $\endgroup$ Commented May 24 at 11:01
  • $\begingroup$ @ten_to_tenth that is incorrect, if $z = 3 + 4i$, $|z\cdot z|=|9-16+24i|\neq 25$. The correct relation is $\Re\left(z\cdot \bar{z}\right)=\vec{OM}\cdot\vec{OM}$ $\endgroup$
    – acat3
    Commented May 24 at 11:06
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You found out that if we have two complex numbers $z_{1}= (a + bi) $ and $z_{2}= (c+di)$ that the product $ac - bd + ( bc +ad)i $ is not what you want for the dot product. You would like $ac + bd$. You want a scalar the goes from negative of the magnitude of the product when they are in opposite directions and to the magnitude of the product when they are in the same direction.

If you want to think of the complex numbers as 2d vectors, where the dot product goes from -(product of lengths) to (product of lengths), what should you do? Take the case of a=c and b =d, and you get you want $(a+ bi)$ with $ ( a + bi)$. Looks like you need the dot product to be defined as

Dot product of $z_{1} ~~and ~~ z_{2} = (a+bi ) (c-di) $.

That is the standard definition of dot product! Taking the minus of the imaginary part (Mirroring vector across the real axis) is called the conjugate. It reverses the rotation of a complex number.

Now if we dot product a complex number by another that does not point in the same direction or in the reverse direction, we still have a problem. You we get a complex number with an imaginary part. What do we do with that? Just throw it away. You get:

Real part of $ {( a+ bi) (c-di )} = ac + bd$,

just like the 2d real dot product. Now taking the real part of a complex expression gets us out of the field of complex numbers, so it shouldn't be done lightly. For complex numbers, products, even using conjugates, do not give real 2d dot products. We have to use the real part of product with conjugate. Taking the real part is not a linear relation over complex numbers. For example, Re($ i\cdot i $) does not equal Re( $i$) $\cdot$ Re($i$), because -1 does not equal zero. So when we have ventured into getting a 2d real dot product, we have gone outside the field of complex numbers. That doesn't mean it is not a useful calculation. We just have to be careful how used.

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