Complex numbers help? (1 Viewer)

[MiseryParade]

"use the vector representation of z1 and z2 on an argand diagram to show that if |z1| = |z2|, then (z1+z2)/(z1-z2) is imaginary."

I have no idea. :/ This is what I tried to do: arg(z1+z2)/arg(z1-z2) = arg (2*z2). I drew the diagram so that z1 and z2 were reflections of each other centred around the imaginary axis, with equal arguments of 45, so that when arg (2*z2), it'd be arg(90), which is on the imaginary axis...

Could someone point me in the right direction? I don't know how to approach questions like these, and there aren't any available answers since it was a "show" question.

Are there any 4U textbooks that have worked solutions to all questions, including proofing?

 

[deswa1]

Consider the parallelogram formed by z1, z2, O and (z1+z2). Now note that |z1|=|z2| so all the side lengths are equal -> therefore its a rhombus

Now z1+z2 is one diagonal and z1-z2 is the other (graph the on your Argand diagram to see). But the diagonals of a rhombus intersect at right angles. Therefore arg(z1+z2)/(z1-z2)=90 and it lies on the imaginary axis etc.

And Terry Lee has full solutions. Cambridge and Fitzpatrick have 'unofficial' answers

 

[manscux]

Hi.. i agree with deswai but this is an alternative

I.e Deswai is right that since |z1|=|z2| OPRQ is a rhombus

Thus OR is perpendicular to QP

Therefore z1+z2 = ki (z1-z2) (diagonals of a rhombus)

than simply divide by z1-z2 and you have shown that it is imaginary

 

[bobmcbob365]

Usually when you see something like z1+z2 and/or z1-z2, you would think about drawing the parallelogram.

If a question asks you to show that something is purely imaginary, you would either: 1. convert into the form x+iy and prove that x =0 or 2. prove that the argument is equal to plus or minus pi/2.

And also, if a question gives you a condition related to the modulus, more often than not, you would need to solve it geometrically, which means probably use the second method. And then you would have to introduce the argument somehow.

 

[Carrotsticks]

Usually when you see something like z1+z2 and/or z1-z2, you would think about drawing the parallelogram.

If a question asks you to show that something is purely imaginary, you would either: 1. convert into the form x+iy and prove that x =0 or 2. prove that the argument is equal to plus or minus pi/2.

And also, if a question gives you a condition related to the modulus, more often than not, you would need to solve it geometrically, which means probably use the second method. And then you would have to introduce the argument somehow.

Very good explanation. There is also manscux's method of showing that Z = ki W, where k is some real non-zero constant.

Now if someone were kind enough to show a diagram as an example....

 

[Arr0gance]

What they said but here's a diagram which may ALSO help you understand it a bit better



NOTE: This solution is "diagram dependent", so it only proves one case. However it will work wherever you draw Z1 and Z2, yielding a result of either pi/2 or -pi/2, which means it is always purely imaginary.

 

[HeroicPandas]

What they said but here's a diagram which may ALSO help you understand it a bit better

View attachment 26634

NOTE: This solution is "diagram dependent", so it only proves one case. However it will work wherever you draw Z1 and Z2, yielding a result of either pi/2 or -pi/2, which means it is always purely imaginary.

I'm curious, what pen do u use?

 

[RealiseNothing]

What they said but here's a diagram which may ALSO help you understand it a bit better

View attachment 26634

NOTE: This solution is "diagram dependent", so it only proves one case. However it will work wherever you draw Z1 and Z2, yielding a result of either pi/2 or -pi/2, which means it is always purely imaginary.

Very nice.

 

[Arr0gance]

I'm curious, what pen do u use?

I think it was one of those promotion pens? Just a normal ball-point pen i think, Why?

 

[HeroicPandas]

I think it was one of those promotion pens? Just a normal ball-point pen i think, Why?

Uh, I like those types of pens, nice and dark, looks smooth too!