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I am trying to solve $$(\cos\alpha-\lambda)^2+\sin^2\alpha=0$$ for $\lambda$. Expanding and using the identity $\sin^2x+\cos^2x=1$ yields $$\lambda^2-2\lambda\cos\alpha+1 = 0$$ and using the quadratic formula gives me $$\lambda=\cos\alpha\pm\sqrt{\cos^2\alpha-1}.$$

The solution to this problem is $$\lambda = \cos\alpha\pm i\sin\alpha$$ but I don't see how to obtain that result.

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  • $\begingroup$ The problem is that you wrote the square root of a negative real, and that is impossible. However if you rewrite $\cos^2{\alpha}-1$ and you solve your equation in $\lambda$ for complex numbers (as you should), you might have a more satisfying result $\endgroup$ – T_O Apr 9 '14 at 14:52
  • $\begingroup$ should it be $\mp i \sin x$? $\endgroup$ – Alex Apr 9 '14 at 14:55
  • $\begingroup$ also note that $\lambda =\cos \alpha \pm i \sin \alpha=e^{\pm i\alpha}$ $\endgroup$ – Jonas Kgomo Apr 9 '14 at 15:06
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Note that $\sqrt{\cos^2 \alpha - 1} = \sqrt{-(1-\cos^2 \alpha)} = \sqrt{- \sin^2\alpha} = i\sqrt{\sin^2\alpha}$...

So $$\lambda=\cos\alpha\pm\sqrt{\cos^2\alpha-1} = \cos \alpha \pm i\sin\alpha$$

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$$\lambda=\cos\alpha\pm\sqrt{\cos^2\alpha-1}=\cos\alpha\pm\sqrt{(-1)(1-\cos^2\alpha)}=$$ $$=\cos\alpha\pm\sqrt{(-1)\sin^2\alpha}=\cos\alpha\pm\sqrt{(-1)}\cdot\sqrt{\sin^2\alpha}=$$ $$=\cos\alpha\pm i\sin\alpha$$

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