Complex roots (1 Viewer)

Utility · Mar 21, 2013

If w₁ = -√3 + i is a cube root of z. Find z. Is there any quick method to getting the original angle (arg(z))? I don’t understand that part, in particular.

Autistic Chimp · Mar 21, 2013

It has been a while since I did complex, but if (-√3+i)^3 = z, then z = 8i when you expand it, therefore arg(z)=pi/2. That's about 4 lines of working.

Utility · Mar 21, 2013

Yeah, pi / 2. I started doubting it after doing so many of them.

RealiseNothing · Mar 21, 2013

$-\sqrt{3}+i$

$2cis(\frac{5\pi}{6})$

Cubing this, using De Moivre's Theorem, we will cube the modulus and multiply the argument by 3:

$2^3cis(\frac{5\pi}{2})$

$8cis(\frac{\pi}{2})$ (note we just subtracted $2\pi$ so it doesn't change anything).

$8i$

Nws m8 · Mar 22, 2013

RealiseNothing said:
$-\sqrt{3}+i$

$2cis(\frac{5\pi}{6})$

Cubing this, using De Moivre's Theorem, we will cube the modulus and multiply the argument by 3:

$2^3cis(\frac{5\pi}{2})$

$8cis(\frac{\pi}{2})$ (note we just subtracted $2\pi$ so it doesn't change anything).

$8i$

That's the right and quickest method, I think they teach you this method at school, don't they? ...

Complex roots (1 Viewer)

Utility

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Autistic Chimp

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Utility

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RealiseNothing

what is that?It is Cowpea

Nws m8

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