How can I solve the complex number equation x^4 + 14 = 0?

[greg997]

Hi there, I ve got problem with this equation.

x^4+ 14 = 0

I tried like this:

X^2 = z

z^2 +14 = 0
z^2 = -14
z= sqr-14
z= j 3.74

then back to x

x^2 = j.374

and now what can i do??

 

[bsdz]

Consider using the roots of unity.

i.e.

$$x^4 - 14 = 0$$

$$x = e^\frac{2 \pi i k}{4} \sqrt[4]{14} ; k = 0, 1, 2, 3$$

That's from memory. You'll need to double check it.

 

[Avodyne]

Both are correct, these answers are the same.

 

[HallsofIvy]

Messed up again! greg997, I meant to click on quote but accidently clicked on "edit".
I hope I have re-established what you had originally.

Hi there, I ve got problem with this equation.

x^4+ 14 = 0

I tried like this:

X^2 = z

z^2 +14 = 0
z^2 = -14
z= sqr-14
z= j 3.74

You should have z=-j sqrt(14).

then back to x

x^2 = j.374

and now what can i do??
The two sqrts of j are $\sqrt{2}/2+ j\sqrt{2}/2$ and $-\sqrt{2}- j\sqrt{2}/2$ and the two sqrts of -j are $-\sqrt{2}/2+ j\sqrt{2}/2$ and $\sqsrt{2}/2- j\sqrt{2}/2$. Multiply those by the fourth root of 14..

The two sqrts of j are $\sqrt{2}/2+ j\sqrt{2}/2$ and $-\sqrt{2}- j\sqrt{2}/2$ and the two sqrts of -j are $-\sqrt{2}/2+ j\sqrt{2}/2$ and $\sqsrt{2}/2- j\sqrt{2}/2$. Multiply those by the fourth root of 14.

How did I get those roots? Well, in the "complex plane", j is at (0,1). The square root will have the same modulus (1) and half the argument: 90 degrees becomes 45 degrees.

 

[Дьявол]

You can do like this:
x⁴=-14
so
$$x=\sqrt[4]{-14}$$

In your case z=-14, so -14=-14+0*i

a=-14 ; b=0

Now $r=\sqrt{(-14)^2+0^2}=14$

$$cos\alpha=\frac{a}{r}$$

$$sin\alpha=\frac{b}{r}$$

At this point you need to find the angle $\alpha$ and substitute in the formula below.

$$w_k=\sqrt[n]{r}(cos\frac{\alpha+2k\pi}{n}+isin\frac{\alpha+2k\pi}{n})$$ for k=0,1,2,...,n-1

 

[greg997]

Thanks everybody for these explanations. They really helped me.