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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