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

CrispyPlanet
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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.
 
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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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