An Interesting Homemade Equation | Problem 459

[@SweetSorrow777]

Yes, 2nd method all the way. Much simpler to write z as r*(e^i* theta) from the beginning.😊

[@scottleung9587]

Cool!

[@FRODOGOOFBALL]

one thing that went wrong in the first method: at 6:30 you said r^2 = 2, but then at 7:00 you said r^2 = 4

[@mcwulf25]

Fell at the last hurdle!

[@aplusbi]

uh-oh! 😁

[@ssalmero]

As 2i=2.e^(pi/2+2npi), the generic argument is theta= (pi/2).(2n+1/2)

[@dixonblog]

z = -(1+i) is also a solution

[@mcwulf25]

Because we only considered arg(2i) = pi/2.
If we did this properly we have arg(2i) = pi/2 + 2n.pi
So after dividing by 2 to find arg(z) we have pi/4 and 5pi/4. (And 9pi/4 etc but these just correspond with the first two).
An argument of 5pi/4 is in q3 and a and b are negative, as you state.

[@mcwulf25]

Method 3
Multiply both sides by complex conjugate:
(z.z*)^(|z|^2) = 2i.(-2i) = 4
(|z|^2)^(|z|^2) = 4 = 2^2
So |z|^2 = 2
|z| = sqrt(2)
z = |z|^2.e^it = 2e^i.pi/2
e^2it = sqrt(2)e^i.pi/2
e^2it = e^i.pi/2
2it = i.pi/2
t = pi/4
z = sqrt(2) e^i.pi/4 = 1 + i

[@Don-Ensley]

problem
z ᑊᶻᑊ^² = 2 i
Let
z = √(a² + b²) e ⁱ ⁽ᵗᵃⁿ⁻¹⁽ᵇᐟᵃ⁾⁺ ᴷ ⷫ ⁾
, where K is an integer. The period of tangent is π.
We have
|z|² = (a² + b²)
2 i = eˡⁿ² ⁺ ⁱ ⷫ ⁽¹ᐟ²⁺²ᴺ⁾
, where N is an integer.
Originally the equation is
z ᑊᶻᑊ^² = 2 i
Take natural logs on both sides.
(a² + b²) { ½ ln(a² + b²) + i [tan⁻¹(b/a)+Kπ] } =
ln 2 + i π (1/2+2N)
Equate Real parts:
(a² + b²) ½ ln(a² + b²) = ln 2
(a² + b²) ln(a² + b²) = 2 ln 2
Equate Imaginary parts:
(a² + b²)[ tan⁻¹(b/a)+Kπ] = π(1/2+2N)
From
(a² + b²) ln(a² + b²) = 2 ln 2
a² + b² = 2
From
(a² + b²)[ tan⁻¹(b/a)+Kπ] = π(1/2+2N)
We show by tangent sums formulas that the result is the same as following the principle branch where N = K = 0.
2 [ tan⁻¹(b/a)+Kπ] = π(1/2+2N)
tan⁻¹(b/a)+Kπ = π(1/2+2N)/2
tan⁻¹(b/a) = π(1/2+2N)/2-Kπ
b/a = tan [(π/4+πN) -Kπ]
= [ tan(π/4+πN)-tan(Kπ) ]/
[ 1 + tan(π/4+πN) • tan (Kπ) ]
tan(π/4+πN) = [tan(π/4)+ tan(πN)]/
[ 1 - tan(π/4)•tan(πN) ]
= [1 + 0] / [ 1 - 0 ]
= 1
b/a = [ 1-tan(Kπ) ]/[ 1 + 1• tan (Kπ) ]
b/a = [ 1-0 ]/[ 1 + 1•0 ]
b/a = 1
a = b
2 a² = 2
a = ± 1
b = ± 1
z = 1 + i, -1-i
answer
z ∈ { -1 - i, 1 + i }