Question

Given the relation e^iz = cosz + isnz and e^-iz = cosz - isinz. Show that
I. 1 + tan^2 z = sec^2 z
2. sin^2 z + cos^2 z = 1

Answer 1

1) First, let us write tangent in terms of exponentials using $\sin x = \frac{e^{ix} - e^{-i x}}{2 i}$, $\cos x = \frac{e^{i x} + e^{-i x}}{2}$:

$\tan x = \frac{\sin x}{\cos x} = \frac{e^{i x} - e^{-i x}}{i(e^{i x} + e^{-i x})}$ .

Using last equation, obtain:$1 + \tan^2 x = 1 - \frac{(e^{i x} - e^{-i x})^2}{(e^{i x} + e^{-i x})^2} = \frac{(e^{i x} - e^{-i x})^2-(e^{i x}+e^{-i x})^2}{(e^{i x} + e^{-i x})^2} = \frac{2 e^{i x} 2e^{-ix}}{4 \cos^2 x} = \frac{1}{\cos^2x} = \sec^2 x$

(Here we used $i^2 = -1$, $(a-b)(a+b) = a^2 - b^2$ .

2) $1 = e^{ix} e^{-i x} = (\cos x + i \sin x)(\cos x - i \sin x) = \sin^2 x + \cos^2 x$