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

[RoughRoad]

In a complex number sum, I have encountered a minute difficulty in understanding a step:

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


Now my question is, how did the 'i' got eliminated from the second step? Now, i equals \sqrt{}-1, so when squared, there should be a minus sign in the second step. Can anyone help me clearing my basics?

 

[Rasalhague]

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:

\left | a \right | = \sqrt{\overline{a}a},

where

a =(\text{Re}(a)+i \, \text{Im}(a)),

\overline{a}=(\text{Re}(a)-i \, \text{Im}(a)),

and Re(a) is the real part of a, and Im(a) the imaginary part.

 

[RoughRoad]

Thanks for the help! Simply ignored this basic rule initially.

 

[micromass]

The equality

|a+bi|=\sqrt{a^2+b^2}

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.

 

[Rasalhague]

As often happens, we have two identities, and whichever is taken as the definition, the other pops out as a theorem.