The complex number system is comprised of real numbers and the imaginary numbers. 3+4i or i√2 or 4 or 0 or -100i or -1+i are examples of complex numbers. ** They can have an imaginary part and also real part (3+4i or -1+i). But they can also be only real (4 or 0) or only imaginary (-100i or i√2). ** We can add/subtract complex number much the same way we add/subtract like terms in algebra. We can multiply them using the distributive property, again much like algebra except we need to recall the **DEFINITION of “i”**, which is what we use to indicate imaginary numbers. **The definition is simple: i ^{2}= -1. **

Notice for Multiplication:

- if z is complex number then z times 1 is z
- if z is complex number then z times i is zi and will be perpendicular to z, that is, zi is found by rotating z 90 degrees counterclockwise.
- if z is complex number then z(c+di). [where c+di is another complex number], will be cz + d(zi). Notice we have z (from step 1 above) multiplied by c and the zi (from step 2 above) multiplied by d.

Let us look at these graphically. Recall the x and y axes are ** perpendicular ( make 90 degree angle**). And from note 2 above, the z and zi are perpendicular. We will see example of this in a moment. Thus the idea was presented to let the y-axis be the “i” axis since z and zi are perpendicular. This gives us a way to graph complex numbers. Look at 3+2i and also -2+3i. I think you will see the

**3+2i has been rotated 90 degrees**in the counterclockwise direction and gives us

**-2+3i.**And from NOTE 2 above, let us

**multiply the 3+2i by i: (3+2i)i = 3i +2i**

^{2}which is 3i +2(-1) or -2+3i.When we multiply by i the new number will be rotated 90 degrees counterclockwise. Look at the graphs of **3+2i and -2+3i.**

You can also use the slopes of the vectors to prove they create a right angle. The product of their slopes will = -1 if they are perpendicular. What is the slope of the orange vector? What is the slope of the green vector? Multiply and you should get -1. Did you?

**To rotate counterclockwise 90 degrees we multiply by i.** Can we find another complex number we can use with multiplication and rotate 30 degrees or 45 degrees rather than the 90 degrees?

Let us look **for** the complex number that rotates us 45 degrees c-cl. Look back at your Unit Circle. We are looking at angles that “move” around a circle?! That is ROTATION.

- angle
**x y** - 30 degrees cos30° sin30°
- 45 degrees cos45° sin45°

And for the complex numbers we have similar table. The **x-column above is the real-part** of the complex number. The **y-column is the imaginary-part**.

Graph (1, 0). Using the UNIT CIRCLE for the complex number 1+0i we will have cos0 +isin0. You know the 1 is the cos0 and the 0 is the sin0.

The cos0° +isin0° does = 1+0i. For complex number 1/2+ i(√3)/2 we have cos60° +i sin60°. And what complex number is represented by cos45° +i sin45°?

What can we multiply by that will rotate 45 degrees rather than 90? It is cos45°+i sin45°.

Another use for rotating is developing the trig identities for sums of angles. This was what the video by 3blue1brown wanted us to see. For **Cos(A+B) we need the rotation of A and then rotation of B.** Recall this is multiplying when we rotate. So what is rotation of A multiplied by rotation of B, of course I mean as a angle or complex number!

**(cosA+i sinA) times ( (cosB+i sinB) = ? ** This is a bit long. If you know the identity already you can use it, but if not then do the multiplying (that is what he did in the video). **I know the identity so I prefer to use it!** but we are learning something new!!

Find the cos (75°) using the idea above. Change it to rotations that we know! Cos75°= cos(45°+30°) : rotate 45° then rotate 30° then what is the cosine value? **Multiply (cos45°+i sin45°) (cos30°+i sin30°) **. USE FOIL or distributive property. We get cos45°cos30° +cos45°(i sin45°) +i sin45°(cos30°) +sin45°sin30° i^{2}. This simplifies to **real part:** cos45°cos30° – sin45°sin30° and **imaginary part:** cos45°(i sin45°) +i sin45°(cos30°). Since we want the** cosine of 75**°, recall the **real part is the cosine** (SEE unit circle above), we will only need simplify the cos45°cos30° – sin45°sin30°.

Below is the complete identity for cos(A+B). We determined this from finding the **“real”** part of the multiplication of (cosA+i sinA)(cosB+i sinB). How could we then find the sin(A+B) ? Let me know.

What is the sin(105°)? What is the cos(105°)? Send me email from Contact page or write in the comments.

We used imaginary numbers to explain trig identities and rotations. Both of these topics are widely used in **science, engineering, computer graphics and animations**. Perhaps imaginary numbers are more real than we think or **maybe they just need a new name! Do you have a suggestion?**

I don’t normally comment but I gotta state thanks for the post on this great one : D.