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Does any $u \in \mathbb{C}$ exist such that:

$$\frac{u}{\sqrt{-u^2}}=1$$

If yes, give an example please.

UPDATE:

OK, I thought a little about that myself and I think it goes like this ($m,n\in\mathbb{N}$ and $r,\phi\in\mathbb{R}$):

$$\frac{u}{\sqrt{-u^2}}=\frac{|r|e^{i(\phi+2\pi n)}}{\sqrt{e^{i(\pi+2\pi m)}(|r|e^{i(\phi+2\pi n)})^2}}=\frac{e^{i(\phi+2\pi n)}}{e^{i\frac{\pi+2\pi m}{2}}e^{i (\phi+2\pi n)}}=e^{-i\pi(\frac{1}{2}+ m)}$$

Since $(\frac{1}{2}+m)$ can never be an even integer, the above equation can never hold.

bollty
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3 Answers3

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No such solution can exist. Squaring both sides gives $$\frac{u^2}{-u^2}=1$$ which can only hold true if $1 = -1$.

guest196883
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  • Could you elaborate a little more on why one apparently can generally assume that $\frac{u}{\sqrt{-u^2}}=\sqrt{\frac{u^2}{-u^2}}$? Your solution is only true if that holds. – bollty Jan 24 '13 at 10:52
  • @bollty We have $a = b \implies a^2 = b^2$, so $(\frac{u}{\sqrt{-u^2}})^2 = 1^2$ is implied by the statement, and this gives my equation without taking the square root of anything. – guest196883 Jan 24 '13 at 10:58
  • Same story, please elaborate on why one generally can assume $(\frac{u}{\sqrt{-u^2}})^2=\frac{(u)^2}{(\sqrt{-u^2})^2}$. I think with complex numbers this step is not trivial. – bollty Jan 24 '13 at 11:07
  • $(\frac{z}{w})^2 = (\frac{z \bar{w}}{w \bar{w}})^2 = \frac{z^2 \bar{w}^2}{w^2 \bar{w}^2} = \frac{z^2}{w^2}$ for complex $z,w$ – guest196883 Jan 24 '13 at 11:34
  • That would again ask for the verification of the $\left(\frac{z^2}{w^2}\right)\cdot\left(\frac{{\bar w}^2}{{\bar w}^2}\right)=\frac{(z^2 {\bar w}^2)}{(w^2{\bar w}^2)}$, which works well in the exponential representation. – bollty Jan 24 '13 at 12:14
  • @bollty and if you were to do that I would recommend you read the first chapter of any book on complex analysis, which proves many of the simple verifications you are asking for such that instead of resorting to long, drawn out, and tedious calculations with cumbersome notation we can think about the actual underlying math. – guest196883 Jan 24 '13 at 12:43
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Proceed like this:

$$u = \sqrt{-u^2}$$

Squaring both sides:

$$u ^ 2 = - u ^2 $$

or $$1 = -1$$

which is contradiction. Hence, not possible.

hjpotter92
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$$\frac{u}{\sqrt{-u^2}}\cdot\frac{u}{\sqrt{-u^2}}=1.1 \implies\frac{u^2}{-u^2}=1 \implies -1=1$$ Your equation only give the solution $1=-1$. So It is not a valid equation whether $u \in \mathbb{C}$ or not.

A.D
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