# Thread: proof with complex numbers

1. ## proof with complex numbers

a, x, y, z are real numbers such that:

$\displaystyle \frac{cosx + cosy + cosz}{cos(x+y+z)}= \frac{sinx + siny + sinz}{sin (x+y+z)} = a$

Using the identity, a = $\displaystyle \frac{cosx + i sinx + cosy + i siny + cosz + i sinz}{cos(x+y+z) + isin(x+y+z)}$

show that:

$\displaystyle a = cos(y+z) + cos(x+z) + cos(x+y)$

thanks

2. Originally Posted by differentiate
a, x, y, z are real numbers such that:

$\displaystyle \frac{cosx + cosy + cosz}{cos(x+y+z)}= \frac{sinx + siny + sinz}{sin (x+y+z)} = a$

Using the identity, a = $\displaystyle \frac{cosx + i sinx + cosy + i siny + cosz + i sinz}{cos(x+y+z) + isin(x+y+z)}$

show that:

$\displaystyle a = cos(y+z) + cos(x+z) + cos(x+y)$
Let $\displaystyle u = e^{ix} = \cos x + i\sin x$, and similarly $\displaystyle v = e^{iy}$ and $\displaystyle w = e^{iz}$. Then u, v, w all have absolute value 1, so their inverses are equal to their complex conjugates: $\displaystyle u^{-1} = \overline{u}$ (and similarly for v and w).

Now show that $\displaystyle a = \frac{u+v+w}{uvw} = v^{-1}w^{-1} + w^{-1}u^{-1} + u^{-1}v^{-1} = \overline{vw+wu+uv}$. But $\displaystyle vw = \cos(y+z) + i\sin(y+z)$, with similar expressions for wu and uv. Deduce that $\displaystyle a = \cos(y+z) + \cos(z+x) + \cos(x+y) - i\bigl(\sin(y+z) + \sin(z+x) + \sin(x)y)\bigr)$. But a, on the left side of that equation, is real, so the imaginary part of the right side of the equation must be zero.