Complex Analysis: Proving Vector z1 Parallel to z2

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SUMMARY

The discussion centers on proving that vector z1 is parallel to vector z2 if and only if Im(z1z2*)=0, where z2* is the complex conjugate of z2. The solution involves converting the vectors into polar form, leading to the equation 0=r1r2(sin Ѳ1cos Ѳ2-cos Ѳ1sin Ѳ2). This simplifies to the sine difference formula, indicating that the angles Ѳ1 and Ѳ2 must be equal for the vectors to be parallel. Thus, the conclusion is that z1 is parallel to z2 when their angles are identical.

PREREQUISITES
  • Understanding of complex numbers and their polar representation
  • Knowledge of trigonometric identities, specifically the sine difference formula
  • Familiarity with complex conjugates and their properties
  • Basic skills in manipulating imaginary components of complex expressions
NEXT STEPS
  • Study the properties of complex numbers in polar form
  • Learn about the geometric interpretation of complex vector parallelism
  • Explore trigonometric identities, focusing on the sine and cosine functions
  • Investigate the implications of complex conjugates in vector analysis
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Students studying complex analysis, mathematicians interested in vector properties, and anyone seeking to understand the relationship between complex numbers and their geometric interpretations.

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Homework Statement


Show that the vector z1 is parallel to z2 if and only if Im(z1z2*)=0

note: z2* is the complement of z2


Homework Equations





The Attempt at a Solution


I would probably convert z to polar form.
so, z1=r1(cos Ѳ1+isin Ѳ1)
z2=r2(cos Ѳ2+isin Ѳ2)
so, z2*=r2(cos Ѳ2-isin Ѳ2)

Then, I would plug it into Im(z1z2*)=0

so, Im(r1(cos Ѳ1+isin Ѳ1)r2(cos Ѳ2-isin Ѳ2))

which is: r1r2sin Ѳ2sin Ѳ2

But I'm not sure where to go from here...

PLEASE HELP!


 
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Your final answer for the imaginary part isn't correct. If you do it right, it might just resemble the trig addition formula for sin(theta1-theta2).
 
Ok, so I recalculated my final value for the imaginary part and got...

0=r1r2(sin Ѳ1cos Ѳ2-cos Ѳ1sin Ѳ2)

So then I got:
sin Ѳ1cos Ѳ2=cos Ѳ1sin Ѳ2

So from here, do I just try to show that Ѳ12 to show that the vectors are parallel?
If so, how would I go about doing that?
 
Doesn't that look like a trig formula for the sine of the difference of two angles?
 

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