Solving Complex Equation: z^4 + 1/2(1 - i3^(1/2)) = 0

[AlanPartr]

Im a first year physics student and am having trouble with this question on complex numbers, any help would be greatly appreciated:

find all the roots of the following equation:

z^4 + 1/2(1 - i3^(1/2)) = 0

I know that z can be expressed as an exponential, but I don't know how or even if it helps. I tried doing it with normal algebra but you get to a stage where you have to find the 4th root of i.

 

[AlanPartr]

just read the post above this, sorry I am in the wrong place

 

[HallsofIvy]

z⁴- 1/2(1 - i3^(1/2)) =
$$z^4- \frac{1}{2}(1- \sqrt{3}i= 0$$
is the same as
$$z^4= \frac{1- \sqrt{3}i}{2}]$$

The best way to solve that is to convert to "polar form":
$$\frac{1}{2}- \frac{\sqrt{3}}{2}$$
becomes
r= 1, θ= π/3 or
$$cos(\frac{\pi}{3})+ i sin(\frac{\pi}{3})$$
or
$$e^{\frac{\pi}{3}}$$

the fourth root of that has r= 1^1/4= 1 and &theta= π/12:
$$cos(\frac{\pi}{12}+ i sin(\frac{\pi}{12}$$
or
$$e^{\frac{\pi}{12}}$$
which, according to my calculator is abpout 0.9659+ 0.2588i.

 

[robert Ihnot]

The cube root of 1 is $$\omega=\frac{-1+\sqrt(-3)}{2}$$ which is what we get when we transpose it opposite z^4.
However, $$\omega =\omega^4.$$ So we need only look at $$\omega$$ times 1, -1, i, -i.

In fact, it can be seen by raising to the fourth power that $$\omega^4-\omega=0.$$