Rotations in the complex plane

Click For Summary
SUMMARY

This discussion focuses on the mathematical concept of rotations in the complex plane using complex numbers. The user accurately describes how multiplying a vector by the imaginary unit i results in an anti-clockwise rotation, while multiplying by -i results in a clockwise rotation. The user also explores complex conjugation, confirming that multiplying a complex number by its conjugate yields a positive real number, specifically Q(3+2i) and Q*(3-2i) resulting in (9+4) = (13,0). The discussion concludes with the affirmation that there is no scenario where this multiplication results in a negative real number.

PREREQUISITES
  • Understanding of complex numbers and their representation in the complex plane
  • Familiarity with the properties of the imaginary unit i
  • Knowledge of complex conjugates and their significance
  • Basic understanding of vector operations in a two-dimensional space
NEXT STEPS
  • Study the geometric interpretation of complex multiplication
  • Learn about the polar form of complex numbers and De Moivre's Theorem
  • Explore the concept of complex modulus and its applications
  • Investigate the relationship between complex conjugates and quadratic equations
USEFUL FOR

Mathematicians, physics students, and anyone interested in understanding the geometric properties of complex numbers and their applications in various fields such as engineering and computer science.

DiracPool
Messages
1,247
Reaction score
514
I'm trying to check my understanding of rotations in the complex plane. Do I have any of this wrong? If so, can you please explain why?

1) Rotations

a) Say we start with a vector, Q, defined on the real number line as (5,0). If I multiply that vector by i, we now have a vector "iQ" that is essentially the same as rotating that vector anti-clockwise 90 degrees whereby the vector Q is now completely on the imaginary number line (0,5).

b) Now say I multiply that vector again by i. At this point I now have a new vector, -Q, (-5,0), since i x iQ=-Q, rotated to the negative portion of the real number line 180 degrees.

c) Now say instead of multiplying the previous iQ by i, I simply square the iQ term. Now I have -Q^2, or (-25,0).

d) If I multiply this -Q^2 again by iQ, I now have an -iQ^3 vector rotated 270 degrees anti-clockwise to (0, -125).

e) Completing the cycle, then, multiplying again by iQ gives me Q^4 takes me back to the positive real number line 360 degrees at (625,0).

Is that correct so far?

2) Now let's look at these rotations clockwise:

a) If I rotate the same vector Q (5,0), clockwise 90 degrees, I get -iQ.

b) If I now square this vector (-iQ), that gives me -Q^2, which I'm guessing amounts to rotating that vector again by 90 degrees clockwise onto the negative real number line (-25,0).

c) Multiplying this -Q^2 term again by -iQ will give me iQ^3 (0,125) a clockwise 270 degree rotation from the original position of the vector.

d) Finally, multiplying again by iQ will rotate the vector once more by 90 degrees to the original positive real number line yielding a vector Q^4 (625,0).

So, in a bit of a detailed manner I'm seeing that, multiplying a vector by i amounts to an anti-clockwise rotation through the complex plane, whereby multiplying that vector by -i amounts to rotating that vector clockwise through the complex plane. Is that correct? And if we progressively square the vector as we move through in either direction, we end up with symmetric spirals running anti-clockwise to the other.

Most importantly, I just want to validate that whether we move clockwise (iQ x iQ) or anti-clockwise (-iQ x -iQ) by 180 degrees, we get the same indistinguishable value of -Q^2. I'm trusting that is the case?

3) Complex conjugation

a) Now say we have 2 conjugate vectors not confined to the number lines per se, but actually in the complex plane, Q (3+2i), and Q*(3-2i). If we multiply these together, my understanding is that we get (9+4) or (13,0), a new vector Q^2 on the positive real number line?

b) If we take these vectors to the other side of the real number line, we have Q (-3+2i) and Q*(-3-2i). Now it looks as though if we multiply these two vectors together we again get (9+4) or (13,0). Is that right? The same as in the 3a example? Is this where the "mod squared" or absolute value squared theme comes in when we talk about multiplying a complex number by its complex conjugate. Why do we need to take the absolute value when it seems to come out positive just with the regular math?

Finally, is there never a case where multiplying a complex number by its conjugate will yield a vector on the negative real number line?

Thanks for your consideration and I apologize if my notation is somewhat off. I'm not a mathemetician :redface:
 
Physics news on Phys.org
DiracPool said:
I'm trying to check my understanding of rotations in the complex plane. Do I have any of this wrong? If so, can you please explain why?

1) Rotations

a) Say we start with a vector, Q, defined on the real number line as (5,0). If I multiply that vector by i, we now have a vector "iQ" that is essentially the same as rotating that vector anti-clockwise 90 degrees whereby the vector Q is now completely on the imaginary number line (0,5).

b) Now say I multiply that vector again by i. At this point I now have a new vector, -Q, (-5,0), since i x iQ=-Q, rotated to the negative portion of the real number line 180 degrees.

c) Now say instead of multiplying the previous iQ by i, I simply square the iQ term. Now I have -Q^2, or (-25,0).

d) If I multiply this -Q^2 again by iQ, I now have an -iQ^3 vector rotated 270 degrees anti-clockwise to (0, -125).

e) Completing the cycle, then, multiplying again by iQ gives me Q^4 takes me back to the positive real number line 360 degrees at (625,0).

Is that correct so far?

2) Now let's look at these rotations clockwise:

a) If I rotate the same vector Q (5,0), clockwise 90 degrees, I get -iQ.

b) If I now square this vector (-iQ), that gives me -Q^2, which I'm guessing amounts to rotating that vector again by 90 degrees clockwise onto the negative real number line (-25,0).

c) Multiplying this -Q^2 term again by -iQ will give me iQ^3 (0,125) a clockwise 270 degree rotation from the original position of the vector.

d) Finally, multiplying again by iQ will rotate the vector once more by 90 degrees to the original positive real number line yielding a vector Q^4 (625,0).

So, in a bit of a detailed manner I'm seeing that, multiplying a vector by i amounts to an anti-clockwise rotation through the complex plane, whereby multiplying that vector by -i amounts to rotating that vector clockwise through the complex plane. Is that correct? And if we progressively square the vector as we move through in either direction, we end up with symmetric spirals running anti-clockwise to the other.

Most importantly, I just want to validate that whether we move clockwise (iQ x iQ) or anti-clockwise (-iQ x -iQ) by 180 degrees, we get the same indistinguishable value of -Q^2. I'm trusting that is the case?

All correct.

3) Complex conjugation

a) Now say we have 2 conjugate vectors not confined to the number lines per se, but actually in the complex plane, Q (3+2i), and Q*(3-2i). If we multiply these together, my understanding is that we get (9+4) or (13,0), a new vector Q^2 on the positive real number line?

Yes.

b) If we take these vectors to the other side of the real number line, we have Q (-3+2i) and Q*(-3-2i). Now it looks as though if we multiply these two vectors together we again get (9+4) or (13,0). Is that right? The same as in the 3a example? Is this where the "mod squared" or absolute value squared theme comes in when we talk about multiplying a complex number by its complex conjugate. Why do we need to take the absolute value when it seems to come out positive just with the regular math?

We don't need to take the absolute value, it just happens to be equal to it. In fact, the absolute value of a complex number ##z## is usually defined by what you did above: ##|z| = \sqrt{zz^*}##.

Finally, is there never a case where multiplying a complex number by its conjugate will yield a vector on the negative real number line?

No, there is never such a case. Prove it.
 
Thanks, good to know I'm on the right track.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K