# 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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