# Can 2 times the dot product of 2 complex numbers equal to this?

• Aug 27th 2012, 07:04 AM
x3bnm
Can 2 times the dot product of 2 complex numbers equal to this?
There's a section in Gibson's "Elementary Geometry of Differentiable Curves:An Undergraduate Introduction" book on page 6:

"Example 1.4: Recall that the component of a vector $\displaystyle a$ in the direction of a unit vector $\displaystyle b$ is the vector $\displaystyle (a \cdot b)b$. (Example 1.1)
It is useful to express this in complex notation. Note that for any vectors $\displaystyle a,b$ we have $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$: in particular when
b is a unit vector (i.e. $\displaystyle b \overline{b} = \left | b \right |^2 = 1$) we have $\displaystyle 2(a \cdot b)b = a + \overline{a} b^2$."

What I don't understand is why and how is: $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$?

The author said(given above) and I quote "for any vectors" this is true.

Can anyone kindly shed light on this and explain a bit why this is so?
• Aug 27th 2012, 07:19 AM
Plato
Re: Can 2 times the dot product of 2 complex numbers equal to this?
Quote:

Originally Posted by x3bnm
"Example 1.4: Recall that the component of a vector $\displaystyle a$ in the direction of a unit vector $\displaystyle b$ is the vector $\displaystyle (a \cdot b)b$. (Example 1.1)
It is useful to express this in complex notation. Note that for any vectors $\displaystyle a,b$ we have $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$: in particular when
b is a unit vector (i.e. $\displaystyle b \overline{b} = \left | b \right |^2 = 1$) we have $\displaystyle 2(a \cdot b)b = a + \overline{a} b^2$."
What I don't understand is why and how is: $\displaystyle 2(a \cdot b) = a \overline{b} + b \overline{a}$?

Suppose that $\displaystyle a=x+iy~\&~b=u+iv$.
You can show that $\displaystyle a\overline{b}+\overline{a}b=2(xu+yv)$.
But as vectors $\displaystyle a\cdot b=xu+yv.$
• Aug 27th 2012, 07:23 AM
Vlasev
Re: Can 2 times the dot product of 2 complex numbers equal to this?
We have

$\displaystyle a\overline{b}+\overline{a}b = a\overline{b}+\overline{a\overline{b}} = 2Re(a\overline{b})$

The real part is just

$\displaystyle a_1b_1-a_2(-b_2) = a_1b_1+a_2b_2 = a\cdot b$
• Aug 27th 2012, 07:34 AM
x3bnm
Re: Can 2 times the dot product of 2 complex numbers equal to this?
Thanks a lot Plato and Vlasev. That makes sense now. Again Thanks.