Solve the equation in cartesian and polar form: x^3 + 4\sqrt{1+i} = 0

[bjgawp]

Homework Statement

Solve: $$x^3 + 4\sqrt{1+i} = 0$$

and express in both cartesian and polar form.

Homework Equations

$$e^{i\theta} = \cos (\theta) + i \sin (\theta)$$

The Attempt at a Solution

What I did was move the constant term to the right hand side and squared both sides to get: $$x^6 = 16 + 16 i$$

which implies: $$x = (16+16i)^{1/6} = \left[16\sqrt{2}\right]^{1/6} e^{\frac{(8k+1)\pi i}{6}}$$

Then I simply sub in k = 0, 1, .., 5 for all my roots. But the original equation is a polynomial of degree 3. There should be only 3 factors. Do I have to test them all to see if they work? Or is there an easier way...

Thanks.

 

[rochfor1]

You introduced these extra roots when you squared both sides. To avoid having to substitute them into the equation, try expressing $$4 \sqrt{ 1 + i }$$ in polar coordinates.

 

[bjgawp]

I don't really see how that helps as it looks fairly complicated.

$$(1+i)^{\frac{1}{2}} = 2^{\frac{1}{4}} e^{\frac{(8k+1)\pi i}{8}}$$

which gives us two forms:
$$z_0 = 2^{\frac{1}{4}} \left(\cos \tfrac{\pi}{8} + i\sin \tfrac{\pi}{8}\right)$$

$$z_1 = 2^{\frac{1}{4}} \left(\cos \tfrac{9\pi}{8} + i\sin \tfrac{9\pi}{8}\right)$$

then substituting each of these back, I get two 3rd degree equations which again would give me 6 roots?

 

[Dick]

I don't really see how that helps as it looks fairly complicated.

$$(1+i)^{\frac{1}{2}} = 2^{\frac{1}{4}} e^{\frac{(8k+1)\pi i}{8}}$$

which gives us two forms:
$$z_0 = 2^{\frac{1}{4}} \left(\cos \tfrac{\pi}{8} + i\sin \tfrac{\pi}{8}\right)$$

$$z_1 = 2^{\frac{1}{4}} \left(\cos \tfrac{9\pi}{8} + i\sin \tfrac{9\pi}{8}\right)$$

then substituting each of these back, I get two 3rd degree equations which again would give me 6 roots?

That's like saying solving x^3+sqrt(1)=0 means solving x^3+1=0 and x^3-1=0. Sure you get 6 roots. The 'sqrt' notation in your notation indicates only one of them. You want the 'principal value' of the square root. It's the pi/8 one.