Is xx 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 ω=Reiθ \omega = R e^{i \theta } where R0 R \geq 0 is real and 0θ<2π 0 \leq \theta < 2 \pi , we define ω=Reiθ/2 \sqrt{ \omega} = \sqrt{R} e^{ i \theta / 2 } .

The math student was asked to determine the value of x=ii 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 : x=ii. x=\sqrt{i}-\sqrt{-i}. Step 2. \hspace{1mm} Squaring both sides, x2=i+(i)2ii. x^2 = i+(-i)-2\sqrt{i}\sqrt{-i}. This might introduce extraneous roots.

Step 3. \hspace{1mm} Simplifying the expression,
x2=02i12×i×i12. x^2 = 0-2 \cdot i ^{\frac12} \times i \times i^{\frac12}. Step 4. x2=2ii\hspace{1mm} x^2 = -2\cdot i \cdot i

Step 5. x2=2\hspace{1mm} x^2= 2

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

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

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