Solving An Equation in Two Ways? | Problem 329

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[NadiehFan]

There is an error at 3:20 where you ignore the term 2i at the right hand side, which invalidates any further steps and so the system in a and b you obtain is incorrect.

At 8:00 you have arrived at

(z + 1)³ = −11 + 2i

which is easy to solve since −11 + 2i = (1 − 2i)³ so we have

(z + 1)³ = (1 − 2i)³

Applying the equivalence A³ = B³ ⟺ A = B ⋁ A = ω₁B ⋁ A = ω₂B where ω₁ = −¹⁄₂ + i·¹⁄₂√3 and ω₂ = −¹⁄₂ − i·¹⁄₂√3 are the complex cube roots of unity we get

z + 1 = 1 − 2i ⋁ z + 1 = (−¹⁄₂ + i·¹⁄₂√3)(1 − 2i) ⋁ z + 1 = (−¹⁄₂ − i·¹⁄₂√3)(1 − 2i)

which gives

z = −2i ⋁ z = (−³⁄₂ + √3) + i·(1 + ¹⁄₂√3) ⋁ z = (−³⁄₂ − √3) + i·(1 − ¹⁄₂√3)

[mcwulf]

I can't see any short cuts either but the modulus of the rhs after cube rooting is sqrt(5). And we have arg(z+1) = -11/2.
z + 1 = sqrt(5) e^(-11/2)i

That's the primary root. Then the other two are at 2pi/3 and 4pi/3 clockwise from it.

[lisaworldpeace]

You stopped too soon. I got the first root, z = -2i, but then it gets tricky

[alexandermorozov]

The cubic root of a complex number should be searched through its modulus and phase, converting the number into an exponential form of the form z=Ae^(i×fi). You will get three answers in increments of 2pi/3 😄 Don't thank me!

[rslitman]

I lost interest in the first method because you forgot to add the 2i in the righthand side.

[pietergeerkens]

Nice problem.

From 8:48, I noted that the norm of -11 + 2i is 125 = sqrt(5) ^ so the primary cube root likely is a simple expression with norm 5; and looking at the signs probably in 2nd or 4th quadrant.

Then a first trial showed that
( -2 + 1 )^3 = -2 + 11i,
from which negating both sides gave
( 1 - 2i )^3 = -11 + 2i
as desired.