Solve $z^4+16=0$ where $z$ is a complex number

Question

The following exercise is related to complex numbers so $z$ is a complex number. Can you please check whether I solved correctly the exercise. $$z^4+16=0$$ $$z^4=16i^2$$ $$z^2=4i$$

I transformed the complex number $4i$ into the trigonometric form, and got:$$4(\cos(\pi +\pi k)+i\sin (\pi+\pi k))$$. So the result is:$$z=2\left[\cos{\left({\pi +\pi k \over 2}\right)}+i\sin\left ({\pi+\pi k\over 2}\right)\right]$$.

The only problem is that in my workbook the result is: $$z=2\left[\cos{\left({\pi +2\pi k \over 4}\right)}+i\sin\left ({\pi+2\pi k\over 4}\right)\right]$$

I hope you'll help me find the mistake. Thank you that you are reflecting over my exercise !

Answer 1

An alternate solution is to factor this way:

$$x^4+16 = (x^2+4)^2 - 8x^2 = (x^2+4+2\sqrt 2 x)(x^2+4-2\sqrt2 x)$$

And solve $x^2+2\sqrt2 x +4=0$ and $x^2-2\sqrt2 x +4=0$. This will give you the exact solutions:

$$x=\pm \sqrt 2 \pm i\sqrt 2$$

Answer 2

$$z^4=-16$$ How $-1=e^{i(\pi+2k\pi)}$: $$z^4=16 \exp\left(i(\pi+2k\pi)\right)$$ $$z=2 \exp\left(\frac{i(\pi+2k\pi)}{4}\right)$$

Answer 3

$$z^4+16=0<=>$$ $$z^4=-16<=>$$ $$z^4=|-16|e^{arg(-16)i}<=>$$ $$z^4=16e^{(\pi +2\pi k)i}<=>$$

(with k is the element of Z)

$$z=\left(16e^{(\pi +2\pi k)i}\right)^{\frac{1}{4}}<=>$$ $$z=2e^{\left(\frac{1}{4}\pi +\frac{1}{2}\pi k\right)i}$$

(with k goes from 0 to 3 -> k=0-3)