why is conjugate of complex number is reflected only in real axis not imaginary axis (ie only imaginary part changes)

Question

conjugate of 7-5i

(7-5i)(7+5i) = 74

 

but 

(-5i+7)(-5i-7) = -74 which is also real

so why is the first case the only definition of the conjugate of a complex number?

 

Answer 1

Not a career mathematician, yet a 40 year teacher, a conjugate results from the changing of the sign between two terms: 3x + 1 and 3x - 1 are conjugates.

The process is not particular to complex numbers. They just follow the rules, as should mathematicians!

Answer 2

Hi Dan,

 

I'm not a career mathematician, so I can only venture a guess at this answer.

 

My guess is, given a complex number z = a + bi, and two different (but perfectly valid) candidates for its conjugate:

 

     z₁ = a - bi

     z₂ = -a + bi

 

We define z₁ to be the conjugate of z simply because z₁ is far more useful (and special) than z₂.

 

 

Let's go back to elementary algebra for a quick example of z₁'s usefulness.

 

Solve for x:

     x² + 2x + 5 = 0

 

We'd use the quadratic formula to find that:

     x = -1 ± 2i

 

The two solutions to this problem are x₁ = -1 + 2i and x₂ = -1 - 2i, the complex conjugate of x₁ (so long as we define the complex conjugate as we currently do -- as z₁, not as z₂).

 

In fact, any time you end up with complex solutions to a quadratic equation, the solutions will always be complex conjugates. We can go further and say that any time you end up with complex roots to any polynomial function (cubic, quartic... quintic(??), etc.), the complex roots will always come in pairs. If 3 + 5i is a root, then 3 - 5i will also end up being a root.

 

z₁ turns out to be pretty useful in this case, while z₂ is kind of useless.

 

There are other examples of the usefulness of z₁ (and the complete uselessness of z₂), and I think you're about to learn all about them in your math class. Complex conjugates, as long as they're defined as z₁, will end up being pretty powerful tools as you continue your studies.

 

I hope this helps. Maybe a real mathematician can pop in and provide you with a more satisfactory answer.  :)