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


If yes, give an example please.


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.

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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 $1 = -1$.

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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. – Sam DeHority 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 $z,w$ – Sam DeHority 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

Proceed like this:

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

Squaring both sides:

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

or $$1 = -1$$

which is contradiction. Hence, not possible.

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Please note, your altering of the shape has the trivial solution $u=0$ which is not true for the initial equation. – bollty Jan 24 '13 at 10:50
Switching sides/cancellation/division by u is applicable only with $u \neq 0$ – hjpotter92 Jan 24 '13 at 10:52

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

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