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.

  • $\begingroup$ 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

Note that $\sqrt{\cos^2 \alpha - 1} = \sqrt{-(1-\cos^2 \alpha)} = \sqrt{- \sin^2\alpha} = i\sqrt{\sin^2\alpha}$...

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.