# Is $x$ real?

Algebra Level 5

A math student was learning about the principle square root function, and extending it to the complex numbers. For a complex number $\omega = R e^{i \theta }$ where $R \geq 0$ is real and $0 \leq \theta < 2 \pi$, we define $\sqrt{ \omega} = \sqrt{R} e^{ i \theta / 2 }$.

The math student was asked to determine the value of $x = \sqrt{i}-\sqrt{-i}$, and wrote down the following steps. Is every step correct? If no, then what is the first incorrect step.

Step 1. $\hspace{1mm}$Consider the complex number $x$: $x=\sqrt{i}-\sqrt{-i}.$ Step 2. $\hspace{1mm}$ Squaring both sides, $x^2 = i+(-i)-2\sqrt{i}\sqrt{-i}.$ This might introduce extraneous roots.

Step 3. $\hspace{1mm}$ Simplifying the expression,
$x^2 = 0-2 \cdot i ^{\frac12} \times i \times i^{\frac12}.$ Step 4. $\hspace{1mm} x^2 = -2\cdot i \cdot i$

Step 5. $\hspace{1mm} x^2= 2$

Step 6. $\hspace{1mm} x = \pm \sqrt{2}$

Step 7. Reject the extraneous root $x = - \sqrt{2}$, and conclude that $x = \sqrt{2}$.

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