eiθ≡cos⁡θ+isin⁡θ This is an IDENTITY, meaning for whatever value of θ.LHS=RHS. Using θ=12π, LHS=ei12π RHS=cos⁡(12π)+isin⁡(12π) =0+i(1) =i

eix2=i⇔eiπ=i⇔eiπ2=i2⇔eiπ=−1⇔eiπ+1=0 which is Euler's identity

eiθ≡cos⁡θ+isin⁡θ This is an IDENTITY, meaning for whatever value of θ.LHS=RHS. Using θ=12π, LHS=ei12π RHS=cos⁡(12π)+isin⁡(12π) =0+i(1) =i

eix2=i⇔eiπ=i⇔eiπ2=i2⇔eiπ=−1⇔eiπ+1=0 which is Euler's identity

Get the solution to "Why does eiπ2=i?"

Helen Lewis

Answered question

2022-01-16

Why does

e^{\frac{i π}{2}} = i ?

Donald Cheek

Beginner2022-01-17Added 41 answers

e^{i θ} \equiv \cos θ + i \sin θ

This is an IDENTITY, meaning for whatever value of

θ . L H S = R H S

.
Using

θ = \frac{1}{2} π

L H S = e^{i \frac{1}{2} π}

R H S = \cos (\frac{1}{2} π) + i \sin (\frac{1}{2} π)

= 0 + i (1)

= i

Ronnie Schechter

Beginner2022-01-18Added 27 answers

Start with Eulers

alenahelenash

Expert2022-01-24Added 556 answers

e^{\frac{i x}{2}} = i \Leftrightarrow \sqrt{e^{i π}} = i \Leftrightarrow {\sqrt{e^{i π}}}^{2} = i^{2} \Leftrightarrow e^{i π} = - 1 \Leftrightarrow e^{i π} + 1 = 0

which is Euler's identity

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