# Is there any way to represent an imaginary number?

Is there any way to represent an imaginary number? Like the square root of -1? Is there any possible way to do this?

Sorry if you think this is a dumb question. I am a 7th grade student in trigonometry. And I just wanted to ask.

The complex numbers $a+bi$ is often represented by the point $(a,b)$ of the ordinary plane. You might want to look at the following brief description, and then at the lengthy Wikipedia article.

So the "imaginary" number $i$ is represented as the point $(0,1)$, Not at all imaginary!

The addition of two complex numbers then has a simple geometric representation. Multiplication of a complex number by another one also has an important geometric representation, essentially a rotation possibly followed by scaling. Multiplication by $i$ turns out to be counterclockwise rotation about the origin through $90$ degrees. So doing it twice is the same as turning through $180$ degrees, which turns $(a,b)$ into $(-a,-b)$, the same as multiplication by $-1$. This is "why" $i^2=-1$.

As you can see, this was a very good question.

• Never thought of it that way Jun 18, 2011 at 4:28
• @Andrew: Complex numbers were first mentioned vaguely around the middle of the 16th century, and began to be fairly commonly used in the 17th century. But it took until the late eighteenth century for the now standard geometric interpretation to be proposed. So it should not be a surprise that you did not think of complex numbers "that way." For a couple of centuries nobody did. Jun 18, 2011 at 5:06

An imaginary number is by definition a number of the form $ai$, where $a$ is a real number and $i$ is a square root of $-1$. The imaginary numbers $ai$ and $-ai$ are the square roots of $-a^2$.

An imaginary number can also be represented in the form $ae^{\pi i/2}$ where $a$ is a real number; this is because $$e^{\pi i/2}=\cos(\tfrac{\pi}{2})+i\sin(\tfrac{\pi}{2})=0+(i\cdot 1)=i$$ (see de Moivre's formula).

If you mean a geometric representation of an imaginary number, then on the complex plane (also called an Argand diagram) we have that the imaginary numbers form the vertical axis (the real numbers are the horizontal axis).

• Hmmm. Interesting I will have to explore more on that. Jun 18, 2011 at 4:22

In electrical engineering, they use complex numbers: the modulus and argument of a complex number represent the amplitude and frequency of a current.

• That, is an interesting fact. Jun 18, 2011 at 15:38

The study of imaginary, and complex, numbers is arguably one of the most beautiful parts of mathematics. If you follow this path, you will learn that complex numbers turn out to be extremely useful in almost all parts of mathematics. What could the square root of $-1$ possibly have in common with the notion of prime numbers? It looks like a strange idea at first. Yet, it turns out that the most interesting unproven hypotheses about prime numbers can be expressed in terms of complex numbers. GEdgar said that electrical engineers use complex numbers. What about chemists, cryptographers, physicists, etc? It's astonishing how fantastically useful this superficially artificial notion is.

,Andrew,

This is a great question for a 7th grader to ask. It is great that you are studying trig at this point. Keep it up!

If you want to understand imaginary and complex numbers it is best to think of them in a two dimensional coordinate plane. You can think of ordinary (real) numbers as being the horizontal axis and imaginary numbers as being the vertical axis. Then we can think of any coordinate pair $(x, y)$ as being represented as a complex number $x + iy$. In some ways these numbers are similar to ordinary numbers (i.e. we can add/subtract them, we can multiply/divide them, we can compute their magnitude) and in some ways they are different (i.e. we cannot sort them from smallest to largest, they have an angle associated with them).

When we add them we merely add the first coordinates (real parts) together and add the second coordinates (imaginary) together to yield the sum. Geometrically, this is the natural extension of adding ordinary numbers on a number line where we stack the numbers up to yield the sum. In the complex plane we do the same thing, except the numbers can point in any direction. We still stack them up, maintaining each complex number's magnitude and direction.

When we multiply complex numbers, it gets really interesting because one complex number will rotate and stretch (or shrink) the second one. It is best (in my opinion) if you think about complex multiplication like this. Let $x = a + ib$ and $y = c+id$. Then think about the product of the two:

$$z = x y$$

as $x$ changing $y$ into $z$. Ordinarily we think of multiplication as the combination of two numbers to form a product (and indeed that is what is happening here), however, if $x$ has a certain magnitude and angle, then multiplying $y$ by $x$ changes $y$ by rotating it by the angle of $x$ and stretching it by the magnitude of $x$. For example, if $x = 1 + i$, then it will rotate any number $y$ by 45 deg in the CCW direction and stretch it by a factor of $\sqrt{2}$.

BTW, if you Google 'complex number' you will get a myriad of resources much better than my feeble explanation.

• A slight suggestion: Could you possibly change "vector" to "complex number" or explain how a complex number is a vector? (I don't think most 7th graders are aware of what a vector is.) I think the latter would be more time-consuming and tangential (but not necessarily in a bad way).
– 000
Jan 22, 2012 at 4:14
• @user22144, thank you for the suggestion. I did want to point out that Andrew's question states that "I am a 7th grade student in trigonometry" so I think he is beyond most 7th graders. Jan 24, 2012 at 4:42
• Hah, yes. I did not mean that as a belittlement of his intelligence or anything like that, I'm sure he--if he isn't aware of what a vector is--is simply unaware due to never being exposed to it. Even some of the most advanced students simply don't know certain things. :)
– 000
Jan 24, 2012 at 21:48
• I agree. I removed the term 'vector' from the answer. Again, thank you for the suggestion. Jan 24, 2012 at 23:38