Solving ODE Problems with e^{2s} + 2e^s +2 = 0

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To solve the equation e^{2s} + 2e^s + 2 = 0, it can be treated as a polynomial by substituting e^s with x, simplifying the problem. The discussion also involves finding the expression λ + wi for a complex number z, specifically for z = -2i. The magnitude |z| is determined to be 2, and the argument arg(z) is found to be -1. However, the user is struggling to derive λ + wi from this information. The conversation highlights the connection between solving ODEs and complex analysis in determining solutions.
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How do I find all the solutions of
e^{2s} + 2e^s +2 = 0
?
 
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That looks like a regular polynomial to me. Replace e^{s}~with~x and see if you can pick it up.
 
okay thanks, i got that.

This is the last part of a problem for which i completed all the other parts. The question asks to find \lambda + wi such that a given z = e^{\lambda + wi}.
For example, given z = 1-i, |z| = \sqrt{2}, arg(z) = -pi/4, so \lambda + wi = \ln{\sqrt{2}} - \frac{\pi}{4}

The last part of the problem asks for the same information, except with z = -2i. I have found |z| = 2, arg(z) = -1, but I can't find out what \lambda + wi is supposed to be.
 
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