Problem With An Exponent | Problem 379

[@scottleung9587]

Cool!

[@trojanleo123]

z = e^(π/4+2πn).e^[i.ln(1/√2)]
Principle value n=0
z = e^(π/4).e^[i.ln(1/√2)]
z = e^(π/4).[cos(ln(1/√2)) + i*sin(ln(1/√2))]

[@0over0]

Let z = re^(iθ). z^i = r^i * (e^(iθ))^i = r^i * e^(-θ). This =
1+i = root(2)e^(i(π/4). So e^(-θ) = root(2), and r^i = e^(i(π/4) (r&θ switch.)
ln(e^(-θ)) = ln(root(2)), θ = -ln(root(2)). And (r^i)^(-i) = (e^(i(π/4))^(-i).
r = e ^ ( (i(π/4)) * (-i) ) = e^(π/4).
z = r*e^(iθ) = e^(π/4) * e ^ (i(-ln(root(2))).