- #1
Matricaria
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Patterns from complex numbers ! URGENT!
- Use de moivre's theorem to obtain solutions for z^n=i for n=3, 4 and 5.
- Generalize and prove your results for z^n=a+bi, where |a+bi|=1.
- What happens when |a+bi|≠1
Relevant Equations:
z^n = r^n cis (n\theta)
r = \sqrt{a^2 + b^2}
\theta = tan^{-1}(\frac{b}{a})
I solved the first bullet:
Here's my solution for n=4 as an example:
z^4 - i = 0
==> z^4 = i
==> z = i^(1/4).
So this is equivalent to trying to find the 4 fourth roots of i.
In polar form, we see that:
i = cos(π/2) + i sin(π/2).
By De Movire's Theorem:
i^(1/4) = cos[(π/2 + 2πk)/4] + i sin[(π/2 + 2πk)/4]
==> i^(1/4) = cos(π/8 + πk/2) + i sin(π/8 + πk/2) for k = 0, 1, 2, and 3.
Thus:
k = 0 ==> z = cos(π/8) + i sin(π/8)
k = 1 ==> z = cos(5π/8) + i sin(5π/8)
k = 2 ==> z = cos(9π/8) + i sin(9π/8)
k = 3 ==> z = cos(13π/8) + i sin(13π/8)
I don't know how to apply generalization in this case..
- Use de moivre's theorem to obtain solutions for z^n=i for n=3, 4 and 5.
- Generalize and prove your results for z^n=a+bi, where |a+bi|=1.
- What happens when |a+bi|≠1
Relevant Equations:
z^n = r^n cis (n\theta)
r = \sqrt{a^2 + b^2}
\theta = tan^{-1}(\frac{b}{a})
I solved the first bullet:
Here's my solution for n=4 as an example:
z^4 - i = 0
==> z^4 = i
==> z = i^(1/4).
So this is equivalent to trying to find the 4 fourth roots of i.
In polar form, we see that:
i = cos(π/2) + i sin(π/2).
By De Movire's Theorem:
i^(1/4) = cos[(π/2 + 2πk)/4] + i sin[(π/2 + 2πk)/4]
==> i^(1/4) = cos(π/8 + πk/2) + i sin(π/8 + πk/2) for k = 0, 1, 2, and 3.
Thus:
k = 0 ==> z = cos(π/8) + i sin(π/8)
k = 1 ==> z = cos(5π/8) + i sin(5π/8)
k = 2 ==> z = cos(9π/8) + i sin(9π/8)
k = 3 ==> z = cos(13π/8) + i sin(13π/8)
I don't know how to apply generalization in this case..