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Mathman23
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Homework Statement
[tex]\{z \in \mathbb{C}| \mathrm{\theta} \in \mathbb{R}||z| = \theta \cdot |z-1|\}[/tex]
Homework Equations
For which real-value of theta is the above no longer a circle?
The Attempt at a Solution
If I choose theta to equal zero then I get [tex]|z| = 0[/tex], but doesn't that still make it a circle? But now the zero circle??
My second attempt:
[tex]\theta \cdot |z-1| \leq r[/tex]
where r is the radius of the circle.
Then I solve this with respect to theta and get
[tex]\theta \leq r \cdot |\frac{1}{z-1}|[/tex]
My third attempt:
since theta is a real number, then [tex]\theta \in [0, \infty \}[/tex]
[tex]|z| = \sum_{\theta = 0}^\infty (\theta \cdot |z-1|) = \mathop{\lim}\limits_{\theta \to t} \int_{0}^{t} \theta |z-1| d\theta = \frac{t^2 \cdot |z-1|}{2}[/tex]
I then insert into the the inequality [tex] |z| = \frac{t^2 \cdot |z-1|}{2} \leq r[/tex]
which can be solved with respect to t:
[tex]|t| \leq \pm \sqrt{\frac{2r}{|z-1|}}[/tex]
Could somebody here please be so kind to look at my attempteds solutions for this problem, and inform me if one or any of them are correct?
Then by inserting I get that [tex] |z| = \frac{(\sqrt{\frac{2r}{|z-1|}})^2 \cdot |z-1|}{2} = r[/tex]
Best regards
Fred
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