Is this question even possible ?

[Guest SNJERIN]

Hi. 

 

I have been trying to solve this question for a while now, but nothing seem to work. In fact, I have started to think that there is something wrong here. 

 

I get that z is pure imaginary only if |z| = 1. So please help me out here. 

[Vioh]

I get that z is pure imaginary only if |z| = 1. So please help me out here. 

 

Not really. "Pure imaginary" means that the complex number must only consist of the imaginary part. In other words, if , then is pure imaginary only if

 

Using this as the starting point, the problem becomes much easier. We basically just have to show that for to satisfy the condition, then must be zero, while can be any real number. The steps are as follow:

1. Let . Substitute this into the equation, we have:
2. By doing a few steps of algebraic manipulations, we will get this polynomial in the end:
3. Solving for , we get:
4. However, since we know that both and must be real, then the only possible solution for is 0, while there's no restrictions whatsoever on what has to be. Hence, we have proven that must be a pure imaginary number

If you have any further question, please feel free to ask

[Guest]

I did it in a different way as I don't get what vioh was saying -- it is me, I am stupid.

 

what I have is -----    I simplified the question into I z + 1 I = I z - 1 I

 

Then expended it into the only 4 cases, which is

 

1, z + 1  = z - 1 ---- cant exist ,

 

2, z + 1 = -z + 1 --- solve it , eventually get z = 0, which means it has only imaginary part.

 

3, -z - 1 = Z - 1 --- solve it, get z = 0, which means it has only imaginary part

 

4, -z + 1 = -z + 1 --- z = 0, cant exist.

 

I hope it makes sense, I cant really explain stuffs but it is just another alternative to solve this question , if I did it correctly.

[maturk]

Hello, I gave it a go also and I'm just wondering if I did it correctly at all. Here was my solution:

 

1. |(z+1)/(z-1)|, replacing z with z=a +bi

 

2. |(a+bi+1)/(a+bi-1)| =1, because taking the modulus of a complex number is the same as taking the magnitude, I go ahead and take the magnitude using the general formula |z|=sqrt(a²+b²)

 

3. Sqrt( ((a+1)²+b²)/((a-1)²+b²))=1, squaring everything and rearranging

 

4. a²+2a+1+b²=a²-2a+1+b², rearranging again

 

5. 4a=0

 

6. a=0

 

Hence for the modulus to hold true, a must be equal to zero which means that the complex number only has an imaginary part. 

Edited March 6, 2015 by maturk