## iop360 Group Title find all complex numbers z satisfying z^4 = -16 in the form a+bi one year ago one year ago

1. ktnguyen1 Group Title

|dw:1353832796610:dw|

2. ktnguyen1 Group Title

|dw:1353833084664:dw|

3. iop360 Group Title

so is that just 2 solutions? shouldnt i have 4?

4. cinar Group Title

|dw:1353825968940:dw|

5. yakeyglee Group Title

It does have four solutions.

6. cinar Group Title

|dw:1353826101375:dw|

7. iop360 Group Title

hmm it says the answers are |dw:1353818990720:dw|

8. yakeyglee Group Title

If you're familiar with complex exponentials, try this: $$\large z = re^{i \theta} \Rightarrow z^4 = r^4 e^{i(4\theta)} = -16 = 16 e^{i\pi}$$ The values that satisfy this are $$\large r=4$$ and $$\large \theta = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}$$. Use $$\large e^{i \theta} = \cos \theta + i \sin \theta$$ to convert to polar.

9. iop360 Group Title

how did you come up with the angles and value for r=4

10. iop360 Group Title

i tried this method originally but im confused at that part

11. iop360 Group Title

$-16 = r^4e ^{i(4\theta)}$

12. yakeyglee Group Title

Sorry...what I meant was $$r = 2$$. We are assuming $$r$$ to be positive and real, and we want it to correspond to a length of 16 since in the complex plane -16 has length (absolute value) |-16|=16. The angles are values $$0\le \theta<2\pi$$ such that $$4\theta \text{ mod } 2\pi = \pi$$.