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- เผยแพร่เมื่อ 27 ม.ค. 2025
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Nice!
You were doing well till the end!!! n = 0, 1 or 2.😂
Of course the "doing it in my head" version was noticing that the rhs has a modulus of 4 and an argument of pi/4. So z^3 has modulus of 16 and argument of pi/2.
z = cbrt(16) * [cube roots of i].
But will there be extraneous solutions because of the sqrt sign in the original equation??? That's something to think about... 😉
problem
z√z = 2√2 + 2√2 i
Let
z = m e ⁱ ᶿ
, where m is the modulus and θ the angle.
The principle square root is then
√z = √m e ⁱ ᶿᐟ²
Replace these in the equation to obtain
(m e ⁱ ᶿ)√m e ⁱ ᶿᐟ² = 2√2 + 2√2 i
The right side is a complex number of modulus
√(8 + 8) = 4
and angle π/4.
m√m e^(3 i θ/2) = 4 e^(i π /4)
We get the system:
m√m = 4
3 i θ / 2 = i π(1/4 + 2N)
, where N is an integer.
m = ∛16
= 2 ∛2
3 θ / 2 = π (1/4 + 2N)
θ = 2 π (1/4 + 2N) / 3
= π (1/6 + 4N/6)
= π ( 4N + 1 ) / 6
z = 2 ∛2 [e^(i π ( 4N + 1 ) / 6 )
for N = 0,1,2
N θ z
0 π / 6 ∛2 ( 1 + i√3)
1 5 π / 6 ∛2 ( -√3 + i )
2 3 π / 2 - 2 i ∛2
For N >= 3 , the angles start to repeat.
answer
z ∈ { -2 ∛2 i,
∛2 (-√3+ i ),
∛2 ( 1 + i√ 3 ) }
z^(3/2) = 2sqrt(2)*(1+i)
z^(3/2) = 4*[1/sqrt(2) + i/sqrt(2)]
z√z=2√2+2√2i
=4(1/√2+i/√2)
=4e^(iπ/4)
∴z=³√16·e(iπ/6)
=³√16(√3/2+1/2·i)
=³√2(√3+i)
why can't u just wait until the END to think about minor roots, i.e,-
z^3 = 16i = 16e^(PI/2),
z = 16^(1/3) * (e^ i [PI/2]) ^ (1/3) = 16^(1/3) * e^ i [PI/6].
Now we add 2nP: z = 16^(1/3) * e^ i [ (PI/6) + 2nPI] ?
Because the 2pi.n also has to be divided by 3
@@mcwulf25 Thanks. I see that.