Fourth power = negative numbers

Number Theory Level 2

What is i \sqrt{i} ?

i = − 1 = − 1 4 \sqrt{i} = \sqrt {\sqrt{-1}} = \sqrt[4]{-1} .

Solve with these formulas ( a b ) 2 = a 2 b 2 (ab)^{2} = a^{2}b^{2} , ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^{2} = a^{2} + 2ab + b^{2} , and ( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^{2} = a^{2} - 2ab + b^{2} .

1 2 ( 1 − i ) \frac{1}{\sqrt{2}}(1-i) 1 2 ( 1 + 2 i ) \frac{1}{\sqrt{2}}(1+2i) 1 2 ( 1 + i ) \frac{1}{\sqrt{2}}(1+i)

. . by . . , 14, South Korea

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1 solution

. .
Feb 4, 2021

Let's check the answers.

Step 1. 1 2 ( 1 − i ) → ( 1 2 ( 1 − i ) ) 2 ⇒ 1 2 × − 2 i = − i \frac{1}{\sqrt{2}}(1-i) \rightarrow (\frac{1}{\sqrt{2}}(1-i))^{2} \Rightarrow \frac{1}{2} \times -2i = -i

Step 2. 1 2 ( 1 + i ) → ( 1 2 ( 1 + i ) ) 2 ⇒ 1 2 × 2 i = i \frac{1}{\sqrt{2}}(1+i) \rightarrow (\frac{1}{\sqrt{2}}(1+i))^{2} \Rightarrow \frac{1}{2} \times 2i = i So, the answer is 1 2 ( 1 + i ) \frac{1}{\sqrt{2}}(1+i) .

Step 3. 1 2 ( 1 + 2 i ) → ( 1 2 ( 1 + 2 i ) ) 2 ⇒ 1 2 × ( − 3 + 4 i ) = − 3 + 4 i 2 \frac{1}{\sqrt{2}}(1+2i) \rightarrow (\frac{1}{\sqrt{2}}(1+2i))^{2} \Rightarrow \frac{1}{2} \times (-3+4i) = \frac{-3+4i}{2}

Since Step 1. and Step 3. are incorrect, so Step 2. is correct.

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