what-the-value-of-i-

Question Number 116822 by bemath last updated on 07/Oct/20

what the value of \sqrt{i} = ?

Commented by Dwaipayan Shikari last updated on 07/Oct/20

\sqrt{i} = (\sqrt{\frac{2 i}{2}}) = \frac{1}{\sqrt{2}} \sqrt{(1 + 2 i + i^{2})} = \pm \frac{1}{\sqrt{2}} (1 + i)

Answered by floor(10²Eta[1]) last updated on 07/Oct/20

\sqrt{i} = \sqrt{e^{\frac{i π}{2}}} = e^{\frac{i π}{4}} = \cos (\frac{π}{4}) + isin (\frac{π}{4})

= \frac{\sqrt{2}}{2} (1 + i)

another solution :

\sqrt{i} = a + bi

i = a^{2} - b^{2} + 2 abi

\Rightarrow a^{2} - b^{2} = 0, 2 abi = i

\Rightarrow a^{2} = b^{2}

\Rightarrow ab = \frac{1}{2} ∴ a^{2} b^{2} = \frac{1}{4}

a^{4} = \frac{1}{4} \Rightarrow a = \frac{\sqrt{2}}{2} = b

\Rightarrow \sqrt{i} = \frac{\sqrt{2}}{2} (1 + i)

Commented by bemath last updated on 07/Oct/20

gave kudos

Answered by 1549442205PVT last updated on 07/Oct/20

i = \cos (\frac{π}{2} + 2 k π) + isin (\frac{π}{2} + 2 k π) = e^{i (\frac{π}{2} + 2 k π)} \Rightarrow \sqrt{i} = {(i)}^{\frac{1}{2}} = e^{i (\frac{π}{4} + k π)} (k = 0, 1)

= \cos (\frac{π}{4} + k π) + isin (\frac{π}{4} + k π) (k = 0, 1)

i) k = 0 \Rightarrow \sqrt{i} = \frac{\sqrt{2}}{2} (1 + i)

ii) k = 1 \Rightarrow \sqrt{i} = - \frac{\sqrt{2}}{4} (1 + i)

Commented by bemath last updated on 07/Oct/20

thank you sir

Commented by 1549442205PVT last updated on 07/Oct/20

Thank Sir . You are welcome

Answered by malwaan last updated on 07/Oct/20

\sqrt{i} = \sqrt{{(\frac{1}{\sqrt{2}})}^{2} + i + {(\frac{i}{\sqrt{2}})}^{2}}

= \sqrt{{(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}^{2}}

= \pm (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \frac{1}{\sqrt{2}} (1 + i)

= \pm \frac{\sqrt{2}}{2} (1 + i)

a n o t h e r m e t h o d

i = [1; \frac{π}{2}]

\Rightarrow \sqrt{i} = [\sqrt{1}; \frac{\frac{π}{2} + 2 k π}{2}]; k = 0; 1

k = 0 \Rightarrow [1; \frac{π}{4}] = c o s \frac{π}{4} + i s i n \frac{π}{4}

= \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} (1 + i)

k = 1 \Rightarrow [1; \frac{5 π}{4}] = c o s \frac{5 π}{4} + i s i n \frac{5 π}{4}

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

∴ \sqrt{i} = \pm \frac{\sqrt{2}}{2} (1 + i)

Commented by bemath last updated on 07/Oct/20

thank you sir

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