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

In summary: Yes, that is correct. By choosing z1=a and z2=bi, we can simplify the problem and show that z1/z2 is purely imaginary. Therefore, the statement is proven and the proof is complete.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 

1. What is a complex number?

A complex number is a number that is composed of a real part and an imaginary part, expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).

2. How do you add or subtract complex numbers?

To add or subtract complex numbers, you simply combine the real parts and the imaginary parts separately. For example, (2 + 3i) + (4 + 2i) = (2 + 4) + (3i + 2i) = 6 + 5i.

3. Can you multiply complex numbers?

Yes, you can multiply complex numbers using the FOIL method. Multiply the first terms, then the outer terms, then the inner terms, and finally the last terms. For example, (2 + 3i)(4 + 2i) = 8 + 4i + 12i + 6i^2 = 8 + 16i + 6(-1) = 8 + 16i - 6 = 2 + 16i.

4. What is the conjugate of a complex number?

The conjugate of a complex number is the same number but with the sign of the imaginary part changed. For example, the conjugate of 2 + 3i is 2 - 3i.

5. How do you divide complex numbers?

To divide complex numbers, you need to multiply both the numerator and the denominator by the conjugate of the denominator. This will result in a real number in the denominator, allowing you to simplify the fraction. For example, (2 + 3i) / (4 + 2i) = (2 + 3i)(4 - 2i) / (4 + 2i)(4 - 2i) = (8 - 6i + 12i - 6i^2) / (16 - 4i^2) = (14 + 6i) / 20 = 7/10 + 3/10i.

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