Is Multiplying a Complex Number by ejθ a Positive Rotation of Its Vector?

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SUMMARY

The multiplication of a complex number \( z \) by \( e^{j\theta} \) results in a positive rotation of its vector by the angle \( \theta \) without changing its length. This is established through the equation \( z(e^{j\theta}) = (a + jb)(\cos\theta + jsin\theta) \), which simplifies to \( (a\cos\theta - b\sin\theta) + (a\sin\theta + b\cos\theta)j \). The original length \( L_0 \) and the new length \( L_1 \) are both equal to \( \sqrt{a^2 + b^2} \), confirming that the length remains unchanged during the rotation.

PREREQUISITES
  • Understanding of complex numbers and their representation in the form \( z = a + jb \)
  • Familiarity with Euler's formula \( e^{j\theta} = \cos\theta + jsin\theta \)
  • Basic knowledge of trigonometry, specifically sine and cosine functions
  • Ability to perform algebraic manipulations involving complex numbers
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  • Study the geometric interpretation of complex number multiplication
  • Learn about the properties of complex conjugates and their role in rotations
  • Explore the application of polar coordinates in complex number analysis
  • Investigate the relationship between complex numbers and linear transformations
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Homework Statement



Hello all, I'm currently reading through "Vibrations and Waves" by A.P French. I'm not very used to these type of books, so I could use a little bit of help right now:

Show that the multiplication of any complex number z by e is describable, in geometrical terms, as a positive rotation through the angle θ of the vector by which z is represented, without any alteration of its length.

Homework Equations



z = a + jb

j = an instruction to perform a counterclockwise rotation of 90 degrees upon whatever it precedes.

cosθ + jsinθ = e

j2 = -1

The Attempt at a Solution



so z(e)
= (a + jb)(cosθ + jsinθ)
= acosθ + (asinθ + bcosθ)j - bsinθ
= (acosθ - bsinθ) + (asinθ + bcosθ)j

Now I'm supposed to show that the above equation is a positive rotation through the angle θ of the vector by which z is represented, without any alteration of its length.

First, I'm going to prove that the length was not altered:

Original length of z, Lo:

Lo2 = a2 + b2
Lo = √(a2+b2)

Length of new z, L1:

L12 = (acosθ - bsinθ)2 + (asinθ + bcosθ)2
L12 = a2 + b2
L1 = √(a2+b2)

Therefore Lo = L1

Now, How do I prove the rotation? I'm not too strong with trig, maybe that's the problem? Can somebody help me out? Thank you!
 
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Have you considered that your complex number can also be described in exponential form using the same relationship:

cosθ + jsinθ = e

if R = |a2 + b2| is the magnitude of the complex number, and ##\phi## = atan(b/a) is the argument of the complex number (equivalent vector's direction angle), then the vector r for the complex number can also be represented by:

r = R(cos##\phi## + jsin##\phi##) = R ej##\phi##

Now go ahead and multiply by your e :wink:
 

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