Olly_s15 Question Thread

[olly_s15]

Answer attached here.

[Mao]

By inspection, I am fairly sure [not 100%] that the answers are wrong, |z-z1|<|z-z2| should give the line Im(z)=-Re(z) [or y=-x], with the bottom left shaded.

Edit: confirmed, the answers are wrong

[Mao]

How to do this question geometrically:

Interpretations:
|z-z1| - distance from z1
|z-z2| - distance from z2

|z-z1| = |z-z2| - the perpendicular bisector of z1 and z2. Note at any point on the perpendicular bisector, distance from the two points are the same.

|z-z1| < |z-z2| - distance from z1 is less than distance from z2. To illustrate this, draw the perpendicular bisector, the side including z1 will satisfy the condition that it is closer to z1 than z2

|z-z1| + |z-z2| = 4 - this is the geometric definition of an ellipse: sum of distances from two points are constant [compare to the definition of a circle: distance from one point is constant]


You can also do this algebraically to convince yourself, but that is too tedious and I can't be bothered.

[olly_s15]

What is the best method to express something like in form?
Can someone please show their working. Thanks.

[brightsky]

A useful way would be to express it as simply: , which helps a lot when your working in cis form. But I'm not 100% sure if this answers your question. Hope this helps.

[olly_s15]

I think i find keeping it in cartesian form and equating real and imaginary parts then solving simultaneously works better.
I think i just answered my own question. Thanks anyway.

[GerrySly]

That's how I did it anyway, allowed me to have it in cis form just in case further parts of the question asked for it. I'm sure theres an easier way I just found it more helpful when I did it this way

[olly_s15]

That is true. Although the question was just simply "express in x + yi form".

[the.watchman]

How about:

Let



So AND









Sub in for b

Is there a shorter way?

[olly_s15]

Well I've been using the same method as the.watchman. But you could use polar form too.

I don't think there is a shorter algebraic way.

[Mao]

Well I've been using the same method as the.watchman. But you could use polar form too.

I don't think there is a shorter algebraic way.

In this case, you'll have to somehow evaluate , which would involve a calculator, or a LOT of trig identities (not undoable, but hard).

You'll also have to take into account the multiple solutions by adding the right increment to the argument. So the algebraic way is the easiest (unless you love trig).

Also, on a side note, denotes the primary square root. The primary root will have the same sign for the imaginary part as original complex number (you can prove it using polar form).

[olly_s15]

new question.. it's been a while

[Mao]

Now, partial fractions.


AHA I see your dilemma. Let me get my thinking cap on.

[Mao]

The integrand is not continuous on the closed interval [-1,1], thus the integral cannot be evaluated directly. Using a limit approach:



So informally, that integral evaluates to negative infinity. Formally, it is divergent (and thus has no integral). It really depends on if you want to take the 'area-under-the-graph' interpretation or not.

[olly_s15]

The integrand is not continuous on the closed interval [-1,1], thus the integral cannot be evaluated directly. Using a limit approach:



So informally, that integral evaluates to negative infinity. Formally, it is divergent (and thus has no integral). It really depends on if you want to take the 'area-under-the-graph' interpretation or not.

Yes this question is quite obscure. The book provides an answer of -pi/2 which is quite strange.

I did what you had done up until you undertook a limit approach and was no good. I think the book has made a typo and a question of this style shouldn't really be in there but meh, when you pay for the book you pay for their mistakes too.

Thanks Mao....