WildTrig15: Complex numbers and rotations

Complex numbers are here explained using geometry and their intimate connection with dilations and rotations. Pure rotations are related to the parametrization of the unit circle.

Category:

Science & Technology

Tags:

• Complex
• numbers
• rational
• trigonometry
• quadrance
• dilations
• rotations
• mathematics
• similarity

[sillypuppy2]

i^a= 1,-1,i, or -i. Is what your trying to say is that squaring these numbers is the same as adding 90 degrees to their bearing?

[njwildberger]

Multiplying a complex number by i rotates it by a quarter turn (90 degrees). So the powers of i are i,-1,-i,1, and then they repeat.

[catalecticant]

A pity you've swept the i^2=-1 issue under the carpet, especially because transformations give you a neat way to introduce i. First, identify dilations with numbers as you have done. Then consider the number -1 as a dilation: it is not the square of another dilation. Then point out that this dilation is also a half-turn: now it *is* the square of something, namely, of a quarter turn in either direction: you can then *define* i as one of them.

[njwildberger]

Perhaps I should have stressed that the rotation(s) of a quarter turn provide natural square roots of the dilation of -1. It is certainly an important point, thanks for bringing it up.

[samruby82]

what would adding rotations look like in this example at @5.56. Are you saying they can't be added at all or that they can't be added and still preserve quadrance?

[jazzzyxjay]

What the heck did he just say? He lost me at hello

[SuperGeetha]

yea yea we knw, no wonder most lunatics are 4m belgium

[heeypollol]

We see this in belgium when we are 14-15 years.

now im 17 and are working on schemes.

This is suposed easy with us?

What do you people see when you're 14-15 'cause i'm interested. Grtz

[mk3v1na]

really informative!!!~

I ACED MY ROTATION TEST, THANKS!!!

ADD ME FOR CYBER

[sillypuppy2]

f(a,b)=x^2 + 2ax + (a^2 + b^2)

Have you ever plugged numbers in for a and b and then tried to find the zeros of the resulting function?

[sillypuppy2]

oh yeah, thx for correcting me. I never went to college but math was my favorite subject in high school and I like to study math.

[7NTM61Ic]

The "multicomplex" class of hypercomplex numbers, with real coefficients, seem to have a directly analogous rotational property to the complex numbers.