The interval between (-2,0) and (2,0), and the entire y-axis (given that pi is the same as -pi)Can someone try this?:
+arg(z+2)=\pi )
I didn't mention a region. I referred to the INTERVAL between (-2,0) and (2,0).
It's pretty easy to visualise if one knows or has seen the answer already, but I think it can be tricky to come up with the answer if one has never seen this type of question before based on just playing with vectors.
I guess it depends on how visual a thinker someone is. For me, I find visual solutions much easier (and much more satisfying) than pure algebraic solutions. And I wouldn't be prepared to label someone a 'maths wiz' until they can visualize these kinds of things.
How'd you get that inequality ?
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Because z^2 -4 is a real negative number it has to be less than 0.
i did that and ekman said that it was the wrong way to do it
maybe i screwed up*Facepalm*is a negative real number, since its arg is
, which means its real part is negative (since negative real numbers have real part < 0), and its imaginary part is 0 (since the imaginary part of any real number is 0).
In hsc, they dont expect you to algebraically prove this locus type q.i did that and ekman said that it was the wrong way to do it
have you done the locus involving arcs of circle?But what if the angle was