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Solve the equation over $\mathbb{C}$ $$z^2-2z+\dfrac{1}{\cos^{2}(\theta)}=0;\quad \theta\in(-\dfrac{\pi}{2};\dfrac{\pi}{2}) $$ and Write solutions in exponential form.

my thoughts:

$\Delta=4-\dfrac{4}{\cos^{2}(\theta)}=-4\tan^{2}(\theta)$ so $\sqrt{-\Delta}=2|\tan(\theta)|$ then

$$z=\frac{2\pm 2i|\tan(\theta)|}{2}=1\pm i|\tan(\theta)|$$

In order to write my solutions in exponential form how I can get rid of the absolute value knowing that theta belongs to the interval $\theta\in(-\dfrac{\pi}{2};\dfrac{\pi}{2}) $

Add update: $$z=\frac{2\pm 2i|\tan(\theta)|}{2}=1\pm i|\tan(\theta)|=\frac{1}{\cos(\theta)}\left(\cos(\theta)\pm i\sin(\theta) \right)=\frac{1}{\cos(\theta)}e^{\pm i \theta}$$

is my proof correct ?

  • 2
    $$\forall t\in\mathbb R\qquad\pm|t|=\pm t$$ – 2017-02-04
  • 0
    @Did yes , thank you – 2017-02-04

0 Answers 0