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Complex Locus

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vikasarkalgud:
Hey Rui,

Still kinda confused about what situations require stuff to be dotted/ not open circle. I dont really get how two 'less than or equal to's' for an argument and circle can create a dotted circle. Thanks for help.

RuiAce:
Can you provide an example? I'm not fully sure I understand what's going on here.

(Essentially you only omit the open circle if equality can always be attained. The instant one equality fails to hold, for example dealing with \(arg (z-z_0) = \theta\) when \(z = z_0\), you have to put an open circle there. Reason why that's the case is because the regions described require both conditions to hold simultaneously, not either or.)

vikasarkalgud:
Sketch the region in the complex plane where the inequalities |z−1| ≤ 2 and
−π/4 ≤ arg(z-1) ≤ π/4  hold simultaneously was the question, and solution had dotted circle

RuiAce:
For that one, there should be an open circle at \(z=1\), since that fails \( -\frac\pi4 \leq \arg(z-1) \leq \frac\pi4 \). However, everywhere else in the required region (including along the sector's boundary) there should certainly not be open circles anywhere.

vikasarkalgud:
yer, got the one at x=1, idk why the solutions had a dotted outer circle, confused me

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