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anonymous
 one year ago
please help
anonymous
 one year ago
please help

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First of all, you know the Argand Diagram is drawn like so.dw:1441202572576:dw Let's do the first part first. The point \(1 + i\sqrt{3}\) has x = 1, y = \(\sqrt{3}\), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To figure out what \(z  1  i\sqrt3 \le 1\) is, let \(z = x + yi\). Substituting in, \(z  1  i\sqrt3 \le 1 \implies (x  1) + (y  \sqrt{3})i \le 1\). Do you think you can simplify this inequality into a form that you are familiar with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a circle with centre \[1+i \sqrt{3}\] ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! So, putting it onto our graph now, dw:1441202964806:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The third part just means that the angle used when graphing is \(\theta \le \frac{\pi}{3}\). We know that \(\theta = \tan^{1}(\frac yx)\). So, we can try to get this into a form we're familiar with. \(\theta \le \frac\pi3\implies \tan^{1}(\frac yx) \le \frac \pi3 \implies \frac yx \le \tan(\frac \pi3) \implies \frac yx \le \sqrt{3} \implies y\le x\sqrt{3}\)Now, we just need to graph this and shade in the right parts. Can you finish it up from there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just want to know what area to shade.. please help me with that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441203537570:dw Just test a point in each section to see whether it works or not. We need to satisfy \((x  1)^2 + (y  \sqrt{3})^2 \le 1\) and \(y \le x\sqrt{3}\). Let's plug in a point above the line but within the circle, like (1, 2). Plugging it in, we can see that it satisfies the circle inequality, but not the line, so we know the shaded area is within the circle. Let's try plugging in a point in the circle, but below the line, like (1, 1). If we plug this in, we can see that it satisfies both inequalities, so we shade that region in.dw:1441203715797:dw
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