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I am came across a problem that asked me to find the magnitude of a complex number:

$$|4 + 3i|$$

After reading online and the textbook, I figured out the question could be solved by finding the distance of the complex number from the origin on a complex plane. As so:

enter image description here

I understand this so far. However, my trouble is that when I solve, using the Pythagorean theorem:

$\begin{align}\sqrt{4^{2} + (3i)^{2}} &= \sqrt{16 + 9i^{2}} \\ &= \sqrt{16 - 9} \\ &= \sqrt{7} \end{align}$

I end up getting the incorrect solution (the correct solution is 5). Now, when I plug $|4 + 3i|$ into my TI-84, I get 5 as a result. I did some research into the magnitudes of absolute values and saw that they can be found by taking the square root of the product of the complex conjugate pair. Solving it this way, I also got 5 as a solution. I am confused as to why the absolute value of a complex number is treated so differently to that of a real number. Why is this and what is the correct way to solve for the magnitude of a complex number?

Edit:

Thanks for all the responses. I'm still having trouble understanding how you can ignore the $i$ when looking at the distance from $0$ to $3i$. I attempted to draw a diagram to better understand this:

enter image description here

I'm rather confused as to why the distance between $0$ and $3$ is $3$, but the distance between $0$ and $3i$ is also $3$. Wouldn't this imply that $|3|$ and $|3i|$ are equal? If so, how is that possible?

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    $\begingroup$ The lengths of the sides are simply $4$ and $3$. There is no $i$ in Pythagora's. $\endgroup$ – dxiv Dec 17 '16 at 23:44
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    $\begingroup$ The modulus of $4 + 3 i$ is the distance in the plane between the points $(0,0)$ and $(4, 3)$. $\endgroup$ – Gribouillis Dec 17 '16 at 23:45
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    $\begingroup$ why the absolute value of a complex number is treated so differently than a real number On the contrary, they are treated exactly the same. You can write any real number $a$ as $a + 0 \cdot i$ and calculate its absolute value as $\sqrt{a^2+0^2}=\sqrt{a^2}=|a|$. $\endgroup$ – dxiv Dec 17 '16 at 23:49
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    $\begingroup$ Draw the picture for |-3| with the points being -3,-2,-1. The distance is 3 not negative 3. If someone asked you how -3 became positive and how can |-3| = 3, what would you answer. ... and yes |3i| = |3| = |-3| = |-3i|. Because absolute values are always positive. i is not positive. It isn't negative. It is a new direction. $\endgroup$ – fleablood Dec 18 '16 at 0:58
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    $\begingroup$ Exactly! The circle also includes $0.93969262078590838405410927732473 +0.34202014332566873304409961468226 i$ and an infinitude of other combinations. All of them a distance of 3 from the origin. $\endgroup$ – fleablood Dec 18 '16 at 1:16
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Well, oh my, you've done it almost right! It is actually given as

$$|4+3i|=\sqrt{4^2+3^2}=\sqrt{25}=5$$

In general,

$$|a+bi|=\sqrt{a^2+b^2}$$

This is because the absolute value only considers $a$ and $b$ as the length of each side of the triangle. Length can't, by definition, be a complex number.

This is like considering the absolute value of a negative number:

$$|-1|=1$$

The absolute value doesn't care which side of $0$ you are at, just how far. So the same goes for complex numbers, and we end up having

$$|3i|=3$$

Now, $|3i|=|3|$ does not mean that $3i=3$, for the reason that it is simply on a different side of $0$. This kind of logic would say $1=-1$, since their absolute values are the same.

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    $\begingroup$ @EricWiener Well, you've defined absolute value as "the distance from $0$", right? And the distance, has nothing to do with complex numbers. It only measures how far from one spot to the next. $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 23:51
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    $\begingroup$ @EricWiener Well, its not the distance from $0$ to $3i$ actually. It's asking you to take a ruler from point a to point b, essentially, distance. $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 23:53
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    $\begingroup$ @EricWiener Intuitively, what is $|3i|$? Draw that, and use a "ruler" to solve. Do you think that would help? $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 23:55
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    $\begingroup$ @EricWiener How about this algebraic approach:$$|x^2|=|x|^2\implies |i^2|=|i|^2$$ $\endgroup$ – Simply Beautiful Art Dec 18 '16 at 0:06
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    $\begingroup$ @EricWiener Oh, better yet:$$|-1|=1\ne-1$$Does this make sense why one would have $|i|=1$? $\endgroup$ – Simply Beautiful Art Dec 18 '16 at 0:08
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The definition (or one possible definition) of the absolute value of the complex number $a+bi$ (where $a$ and $b$ are real) is $\sqrt{a^2+b^2}$. So you ask, why is it not $\sqrt{a^2+(bi)^2}$ instead? The answer is that this is simply how we choose to define it. You could define a different quantity which is $\sqrt{a^2+(bi)^2}$, but you would have to give a different name to it because everyone else has already agreed that "absolute value" means to take $\sqrt{a^2+b^2}$ instead.

Now, a more interesting question is why everyone else decided on that definition. One reason is that you can represent complex numbers as points in the plane by letting $a+bi$ correspond to the point $(a,b)$, and then $|a+bi|$ is the distance from this point to the origin. Note that when you do this, $(a,b)$ is just a point in the ordinary Euclidean plane: it is a point that is $a$ units to the right of the origin and $b$ units above the origin. It doesn't make sense to say that the vertical distance is $bi$, since in geometry when we measure distances they are always positive real numbers. The vertical distance is $b$ because you have moved $b$ units vertically in the plane. (Actually, this is only accurate if $b$ is positive: if $b$ is negative, you have moved $-b$ units down, and so the distance is $-b$ rather than $b$. But you end up squaring this quantity when you use the Pythagorean theorem, so it doesn't matter if it's negative.)

Now ultimately this explanation is not very satisfying, because it doesn't explain why we chose to represent complex numbers in the plane this way. For instance, why don't we choose to represent them such that the complex number $i$ corresponds to a vertical distance different from $1$, or a distance in some direction other than vertical? One answer is that choosing $i$ to mean "go up one unit" happens to make distance have nice algebraic properties we would like absolute values to have. For instance, for real numbers, it is true that $|xy|=|x||y|$. Defining absolute values of complex numbers by $|a+bi|=\sqrt{a^2+b^2}$, it turns out that this is true for complex numbers as well. As a simple example, if we want $|xy|=|x||y|$ to be true for complex numbers, then we should have $|i|^2=|i^2|=|{-1}|=1$. So we should define $|i|=1$ or $-1$, and it is sensible to define it to be $1$ instead of $-1$ since absolute values are supposed to be positive.

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  • $\begingroup$ A good answer, though late, is still a good answer. +1 $\endgroup$ – Simply Beautiful Art Dec 18 '16 at 0:20
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You ask:

Wouldn't the distance from $0$ to $3i$ be $3i$, not $3$?

No, not quite, for the same reason that the distance from $3$ to $0$ is not $-3$. Distance is not simply the difference, but the absolute value of the difference.

Of course, you'll probably find this unhelpful and circular. There are a few different ways you can think about $|z|$:

  • Length of the line from $0$ to $z$ (note that lengths are always positive, no matter if along the positive reals, negative reals, or any other direction of the complex plane)
  • $\sqrt{z\bar z}$ (multiplying by the complex conjugate $\bar z$ takes any number to the positive reals - it's equivalent to the $\sqrt {x^2}$ formula from the real numbers)
  • the radius $r$ when the number is expressed as $r(\cos\theta+i\sin\theta)$, also known as polar form
  • the nonnegative real number you "land on" if you rotate the plane around the origin so your point ends up on the nonnegative part of the real axis

But all of these are restricted to being positive.


If complex absolute value was defined as $\sqrt{a^2 + (bi)^2}$, then starting at $z=3$ and "moving" upwards along the positive imaginary direction would decrease the absolute value. If we want absolute value to measure "distance from $0$" in some sense, then moving farther away from $0$ would make $\sqrt{a^2 + (bi)^2}$ a smaller number! We want our absolute value to increase when we move away from $0$, not decrease.

We want our definition of "absolute value" to not change if we rotate the plane around the origin, and we want the absolute value of a nonnegative real number to equal itself. These make our only option for the definition of $|z|$ to be $\sqrt{a^2 + b^2}$.

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When dealing with a space of numbers, such as a one-dimensional real line $\mathbb{R}$, a two dimensional real space $\mathbb{R}^2$, or the complex plane $\mathbb{C}$, we often need a sort of measure to compare the individual elements. This measuring apparatus is called a norm. So far, you've been exposed to the Absolute Value and Euclidean Norms. The whole purpose of a norm is to assign a positive value to each non-zero number in the space (vectors, really, but let's not be picky) and the value of zero to, well, the zero element.

For real numbers on the one-dimensional number line $\mathbb{R}$, it is sensible to assign the size as the absolute value, $|x|=\sqrt{x^2}$, which gives us the distance of a number from zero.

real line

In a two-dimensional real space, such a plane where we have ordered pairs $(x,y)$ assigned to every point, an intuitive norm that coincides with our physical sense of distance is given by the Pythagorean theorem. $$||(x,y)|| = \sqrt{x^2+y^2}$$ This is the Euclidean norm. It is not the only norm we could equip $\mathbb{R}^2$ with, but it is the most frequently used one. Coincidentally, this Euclidean norm extends for usage on any $n$-dimensional real space, $$||(x_1,x_2,\dots,x_n)|| = \sqrt{x^2+x_2^2+\dots+x_n^2}$$ and, loosely speaking, gives the distance of the number from the origin.

real2

For complex numbers, we also want to create a norm. Again, we want some apparatus to compare the sizes of numbers. By definition, we want this to be a mathematical device that takes in any non-zero complex number $z=x+iy$ and returns a positive real number. The Pythagorean formula almost provides this, but as you saw, it would return a complex number. That's no good. Our solution is to purposefully define the Euclidean norm for a complex plane as

$$|z| = |x+iy| = \sqrt{(x+iy)(x-iy)} = \sqrt{x^2 + y^2}$$

Note that we take in a complex number, but return a non-negative, real one. That's critical. There is no notion of order for complex numbers (is $i$ greater than $1$, less than, or equal to?), yet by returning a real number, we now are able to compare sizes.

Note, now, how similar it is to compare the magnitudes of numbers on $\mathbb{R}^2$ and $\mathbb{C}$.

complex

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$$|z|^2=z\bar z\implies |4+3i|^2=(4+3i)(4-3i)=16+9=25\implies |4+3i|=5.$$

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    $\begingroup$ I think that'll require more explanation to the OP. $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 23:46
  • $\begingroup$ Why can (4+3i)^2 be represented at (4+3i)(4-3i)? $\endgroup$ – Eric Wiener Dec 18 '16 at 0:06
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    $\begingroup$ In my opinion this avoids the issue and heart of the OP's question if absolute value means distance from origin how do you calculate it. If we simply say "absolute value means something that looks entirely different and which has nothing apparently to do with distances at all" we aren't really explaining anything. Of course $\sqrt{z\overline z}$ has everything to do with distance. But it's far from apparent. $\endgroup$ – fleablood Dec 18 '16 at 0:49
  • $\begingroup$ @EricWiener $(4+3i)^2 = (4+3i)(4+3i)$. However $|4+3i|^2 = (4+3i)(4-3i)$ $\endgroup$ – zahbaz Dec 18 '16 at 18:44
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You are correct when you say that

$\vert \vert x\vert \vert ^2 =x \bar x $,

where $\vert \vert x\vert \vert ^2$ is the magnitude squared and $\bar x $ is the complex conjugate.

It is not too hard to show that if $x=a+bi$,

$\vert \vert x\vert \vert ^2 = a^2 + b^2$.

The only error in your calculation and in your picture is the inclusion of $i$ in your altitude of the right triangle, which caused a -9 as opposed to 9.

Note that the absolute value of a real number is just that number's distance to the origin (we take 0 to be the origin for a real number). The complex version of the absolute value, the magnitude, actually tries to do the same thing$-$it gives the distance from a complex number to the origin.

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    $\begingroup$ the absolute value of a real number is just that number's distance to the origin (which is 0 for a real number) It's not $0$, you must have meant something else. $\endgroup$ – dxiv Dec 18 '16 at 0:15
  • $\begingroup$ @dxiv Andy is referring to the origin as $0$, but the wording makes it seem like he's referring the the distance as $0$. $\endgroup$ – zahbaz Dec 18 '16 at 19:37
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    $\begingroup$ @zahbaz Quite possibly, but the origin is one and the same for all numbers in the complex plane, so the bolded part in "distance to the origin (which is 0 for a real number)" makes it even more confusing. I wish Andy clarified the wording. $\endgroup$ – dxiv Dec 18 '16 at 19:44

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