Search results

@ISAM77... have you noticed this question / thread, as it touches on what we were discussing. Taking the question as written, the locus of z where \arg{\left(\frac{z + 2}{\sqrt{3} + i}\right)} \ge \frac{\pi}{12} \qquad \qquad \text{and} \qquad \qquad |z| \le 2 is actually not the answer given...

Perpendicular distance from z = -i to the interval... I get it as 3\sqrt{2} / 2 = \sqrt{4.5}.

I'm assuming that the z must lie in the region established earlier in the question. If it was to the region where |z| \le 2, the minimum value of |z + i| would be zero, occurring when z = -i.

Oops x 2, that's wrong too. The closest point in the locus will be on the interval from z = -2 to z = 2i where the line to z = -i is perpendicular to the interval. I think this gives \frac{3\sqrt{2}}{2} \le |z + i| \le 3

Oops, you wanted min... So, you have: \sqrt{5} < |z + i| \le 3

So, @astj, the third part is asking you to find the point in the locus that is furthest from -i, as |z + i| = |z - (-i)|, which refers to the length of the vector from -i to z. This point is z = 2i, so the answer is \text{max}(|z+i|) = |2i + i| = 3.

Correct but a little careless, in that the circle and line cross at exactly z = 2i.

\begin{align*} \arg{\left(\frac{z + 2}{\sqrt{3} + i}\right)} &\ge \frac{\pi}{12} \\ \arg{(z + 2)} - \arg{\left(\sqrt{3} + i\right)} &\ge \frac{\pi}{12} \\ \arg{(z + 2)} - \frac{\pi}{6} &\ge \frac{\pi}{12} \qquad \text{as $\arg{\left(\sqrt{3} + i\right)} =...

\text{Arg}(z - i) = \text{Arg}(z) - \pi is, indeed, one possible correct answer. Two other possible answers are \arg{(z - i)} = \arg{(z)} - \pi and \arg{(z - i)} = \arg{(z)} + \pi The difference is the difference in the meaning of "arg" and "Arg". Both \arg{z} and \arg{(z)} refer to the...

Two ways to make exam questions difficult are to make the content in the topic challenging and to make you work across topics. The enrichment questions are generally along the lines of the first way. The more content you cover, the more scope there is for the second way.

There is always a purely algebraic method, which is usually awful. There are sometimes purely geometric methods (like for arg(z - i) = arg(z + 1) etc.). If there isn't an obvious purely geometric approach, the efficient answer is likely to involve: treating the complex numbers as vectors...

How about this... if I help you see why your answer is wrong, maybe you can see what an answer should be... (I say "an" answer as there are more than one possible equations...) You have said that the equation for the locus of \left\{z = iy \in \mathbb{C}:\ 0 < y < 1,\ y \in \mathbb{R}\right\}...

3/4... (A) is not correct.

A related question... In addition to figuring out the answer, you might give equations for each of the graphs shown.

Oops... your question was \arg{(z + i)} = \arg{(z - 1)}. I have done a different question. You are right that one of the issues is in quadrants, because inverse tan can only return a value between -pi/2 and pi/2. So, with that restriction recognised, in my working, the LHS is restricted to x...

University websites are under the control of the faculty and university administrations. Some of them have little insight into the student experience and how things might appear / operate from any perspective other than their perspective as administrators... but you often won't be lucky enough...

That algebra is not correct. However, I agree with you that this sort of problem is much easier to do geometrically. The problem with algebraic solutions to problems like these is spotting the limitations / exclusions. Consider, for example, the following working for this problem, which does...

Interpreting \arg{(z - i)} = \arg{(z + 1)} in vector terms, the equation means: the direction of the vector from i to z is the same as the direction from -1 to z This includes every point on the line through i and -1 EXCEPT FOR: (a) those points themselves, as they lead to a term \arg{0}...

\begin{align*} \text{RHS} &= \frac{1}{2}\left(\left|\overrightarrow{AB}\right|^2 + \left|\overrightarrow{AC}\right|^2\right) \\ &= \frac{1}{2}\left(\left|\overrightarrow{a}\right|^2 + \left|\overrightarrow{b}\right|^2\right) \\ &= \frac{1}{2}\left(\overrightarrow{a} \cdot \overrightarrow{a} +...

I agree with @liamkk112, use the fact that the magnitude of a vector, squared, is equal to its dot product. This question is a variation on a standard question covered in the syllabus, that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four...