Proving that z1/z2 is purely imaginary: A Complex Number Problem

Click For Summary

Homework Help Overview

The discussion revolves around proving that the quotient of two complex numbers, z1 and z2, is purely imaginary under the condition that the magnitudes of their sum and difference are equal. The problem involves exploring the properties of complex numbers and their geometric interpretations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various representations of the complex numbers and the implications of the condition |z1+z2| = |z1-z2|. There are suggestions to square both sides of the equation and to explore the relationships between the components a, b, c, and d. Some participants consider geometric interpretations in the complex plane.

Discussion Status

The discussion is active, with participants offering different perspectives and hints. Some guidance has been provided regarding algebraic manipulation and geometric considerations, but there is no explicit consensus on a single approach or solution yet.

Contextual Notes

Participants note the challenge of finding appropriate values for z1 and z2 that satisfy the given condition, as well as the need for a deeper understanding of the geometric implications in the complex plane.

CrispyPlanet
Messages
2
Reaction score
0
Two complex numbers z1 and z2 are taken such that |z1+z2|=|z1-z2|, and z2 not equal to zero.
Prove that z1/z2 is purely imaginary (has no real parts).

I started by taking z1=a+bi, and z2=c+di, then z1+z2=a+c+i(b+d) and z1-z2=a-c+i(b-d)

|z1+z2|=√(a+c)^2 + (b+d)^2
|z1-z2|=√(a-c)^2 + (b-d)^2

As these don't equal each other, I need to choose another two complex numbers. But I'm not sure which ones to choose. I'm assuming this is the hardest part of the problem, and having found z1 and z2 I can divide them to show that no real parts remain. Any help is much appreciated, and thank you in advance.
 
Last edited by a moderator:
Physics news on Phys.org
I think you're confused - the point is that these typically aren't equal to each other, but if they accidentally happen to be, then you get some additional conditions on a, b, c and d that will let you prove that z1/z2 is purely imaginary.

For starters, I would recommend squaring both sides. Then the condition |z1+z2| = |z1-z2| can be re-written as
(a+c)2 + (b+d)2 = (a-c)2 + (b-d)2.

Before you start doing algebra, it might help to figure out what the relationship between a, b, c and d are that will make z1/z2 purely imaginary.

Also, I would recommend drawing a picture for this. In the complex plane, put a dot for z1. What are the (geometric) conditions for z2 that will make |z1+z2| = |z1 - z2|,, what are the geometric conditions for z1/z2 to be purely imaginary? This requires a reasonable amount of knowledge/familiarity with geometry in the complex plane so these might be difficult to answer.
 
CrispyPlanet said:
Two complex numbers z1 and z2 are taken such that |z1+z2|=|z1-z2|, and z2 not equal to zero.
Prove that z1/z2 is purely imaginary (has no real parts).

I started by taking z1=a+bi, and z2=c+di, then z1+z2=a+c+i(b+d) and z1-z2=a-c+i(b-d)

|z1+z2|=√(a+c)^2 + (b+d)^2
|z1-z2|=√(a-c)^2 + (b-d)^2

As these don't equal each other
But what conditions on a, b, c, and d make it so that they are equal? You don't need two more complex numbers.
CrispyPlanet said:
, I need to choose another two complex numbers. But I'm not sure which ones to choose. I'm assuming this is the hardest part of the problem, and having found z1 and z2 I can divide them to show that no real parts remain. Any help is much appreciated, and thank you in advance.
 
There is a nice division which simplifies the problem significantly, and the geometric interpretation can help to find that.

Hint: how would you prove it if z1 was known to be real?
 
Hi Office_Shredder,

If I make z1=a and z2=bi, then both the sum and the difference of both look like a complex number and its conjugate respectively. Also, |z1+z2|=|z1-z2| .

z1/z2 then becomes a/bi. Multiplying top and bottom by bi produces -(a/b)i, which is only imaginary.

Is this correct?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
6K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
4
Views
2K