Why does the 'i' disappear in the simplification of a complex number sum?

  • Level: Undergrad 
  • Thread starter Thread starter RoughRoad
  • Start date Start date
  • Tags Tags
    Complex Complex number
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 3K views
RoughRoad
Messages
63
Reaction score
0
In a complex number sum, I have encountered a minute difficulty in understanding a step:

[itex]\left|(cos\theta-1)+i.sin\theta\right|[/itex]
= [itex]\sqrt{}(cos\theta-1)^2+sin^2\theta[/itex]


Now my question is, how did the 'i' got eliminated from the second step? Now, i equals [itex]\sqrt{}-1[/itex], so when squared, there should be a minus sign in the second step. Can anyone help me clearing my basics?
 
Physics news on Phys.org
Are you remembering that the modulus of a complex number is the square root of the product of the number with its conjugate? That is:

[tex]\left | a \right | = \sqrt{\overline{a}a},[/tex]

where

[tex]a =(\text{Re}(a)+i \, \text{Im}(a)),[/tex]

[tex]\overline{a}=(\text{Re}(a)-i \, \text{Im}(a)),[/tex]

and Re(a) is the real part of a, and Im(a) the imaginary part.
 
Thanks for the help! Simply ignored this basic rule initially.
 
The equality

[tex]|a+bi|=\sqrt{a^2+b^2}[/tex]

is just the definition of the absolute value! There is no reasoning behind it, it's just true by definition. Your OP was also true by definition.
 
As often happens, we have two identities, and whichever is taken as the definition, the other pops out as a theorem.