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

The math student was asked to determine the value of x = i − − i 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 = i − − i . x=\sqrt{i}-\sqrt{-i}. Step 2. \hspace{1mm} Squaring both sides, x 2 = i + ( − i ) − 2 i − i . 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 ⋅ i 1 2 × i × i 1 2 . x^2 = 0-2 \cdot i ^{\frac12} \times i \times i^{\frac12}. Step 4. x 2 = − 2 ⋅ i ⋅ i \hspace{1mm} x^2 = -2\cdot i \cdot i

Step 5. x 2 = 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} .

Step 1 Step 2 Step 3 Step 4 Step 5 Everything is correct

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Sabhrant Sachan by Sabhrant Sachan , 21, India

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2 solutions

Chung Kevin
Dec 28, 2016

There is no error, other than potentially introducing an extraneous root in step 2, which has been accounted for.

With the principle square root function, there is no longer any ambiguity in what these terms are. Let's use the exponential form for these complex numbers to rewrite the steps.

Step 1: x = e i π / 4 − e i 3 π / 4 x = e^{ i \pi / 4 } - e^{ i 3 \pi / 4 }
Step 2: x 2 = e i π / 4 − 2 e i π / 4 e 3 i π / 4 + e 3 i π / 2 x^2 = e^{ i \pi /4 } - 2 e^{ i \pi / 4 }e^{ 3i \pi / 4 }+e^{ 3i \pi / 2 }
Step 3: x 2 = − 2 e i π / 4 e i π / 2 e i π / 4 x^2 = - 2 e^{ i \pi / 4 } e^{ i \pi / 2 } e^{ i \pi / 4 }
Step 4: x 2 = − 2 e i π / 2 e i π / 2 x^2 = -2 e^{ i \pi / 2 } e^{ i \pi / 2 }
Step 5: x 2 = 2 x^2 = 2
Step 6: x = ± 2 x = \pm \sqrt{2}
Step 7: At this point, we have to check back against the original expression. Since the real part of e i π / 4 e^{ i \pi / 4 } is positive and the real part of − e 3 i π / 4 -e^{ 3 i \pi / 4 } is also positive, we conclude that the real part of x x is positive so x = 2 x = \sqrt{2} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

We know that − 4 ∗ − 9 ≠ ( − 4 ) ( − 9 ) \sqrt{-4} * \sqrt{-9} \neq\ \sqrt{(-4)(-9)} . So I don't think the step 4 is correct where i 1 2 ∗ i 1 2 i^\frac{1}{2}*i^\frac{1}{2} is written simply as i. If you cannot multiply square roots of negative numbers, you cannot multiple 4th roots of negative numbers(square root of square root of -1).

Ajinkya Shivashankar - 4 years, 5 months ago

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You are correct to say that we have to be careful with complex exponentiation. The safest way is to use the principal value of the square root.

For example, − 4 × − 9 = ( 2 e i π / 2 ) × ( 3 e i π / 2 ) = 6 e i π = − 6 \sqrt{ -4 } \times \sqrt{ - 9 } = ( 2 e ^ { i \pi / 2 } ) \times ( 3 e ^ { i \pi / 2 } ) = 6 e^{i \pi } = -6 , whereas ( − 4 ) × ( − 9 = 36 = 6 e 0 = 6 \sqrt{ ( -4) \times (-9} = \sqrt{ 36 } = 6 e ^{ 0 } = 6 .

In the step 2, 3, 4 above, we see that the complex exponentiation steps are legally done.

Calvin Lin Staff - 4 years, 5 months ago

I still not get error

ankit raj - 4 years, 5 months ago

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You are correct, there is no error, except maybe introducing an extraneous root in step 2. @Calvin Lin can you fix this problem?

Chung Kevin - 4 years, 5 months ago

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The question is a little confusing , but the final answer is definitely wrong .

i = ± 1 2 ( 1 + i ) − i = ± 1 2 ( 1 − i ) \begin{aligned} \sqrt{i} & = \pm \dfrac{1}{\sqrt{2}}\left( 1+i \right) \\ \sqrt{-i} & = \pm \dfrac{1}{\sqrt{2}}\left( 1-i \right) \end{aligned}

Subtract the two equations

i − − i = ± 1 2 [ ( 1 + i ) − ( 1 − i ) ] = ± 2 i x = ± 2 i \sqrt{i}-\sqrt{-i} = \pm \dfrac{1}{\sqrt{2}}\left[ (1+i)-(1-i)\right] = \pm \sqrt{2} i \\ x = \pm \sqrt{2} i

Sabhrant Sachan - 4 years, 5 months ago

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@Sabhrant Sachan – I disagree. You chose the wrong signage. What you are saying is similar to saying that i + i = 0 i + i = 0 because we make the first i i the principle (positive) root and make the second i i the other (negative) root. Unfortunately, we can't choose the signage arbitrarily.

In this case, the correct equation is: i − − i = 1 2 ( 1 + i ) − ( − ( 1 2 ( 1 − i ) ) ) = 2 \sqrt{ i } - \sqrt{ -i} = \dfrac{1}{\sqrt{2}}\left( 1+i \right) - ( - (\dfrac{1}{\sqrt{2}}\left( 1-i \right)) ) = \sqrt{2} .

Note: Your calculation is valid if the principle square root branch that is chosen is − π / 2 < θ < π / 2 - \pi / 2 < \theta < \pi / 2 . This is not commonly used, which is why I defined it in the question.

Calvin Lin Staff - 4 years, 5 months ago

In step 3, that should be i^½*(-i)^½ which means i^0=1 Then, x² = -2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you provide a more complete reasoning for how we introduced that error? What was the mistake that we made?

Calvin Lin Staff - 4 years, 5 months ago

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