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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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