# Complex numbers

i am to find the 4th roots of -16

$$(-16)^{1/4}=2i^{1/4}$$
$$i=e^{i \pi/2}$$
$$i^{1/4}=e^{i \pi/8}$$
$$(-16)^{1/4}=2e^{i \pi/8}$$
or
$$(-16)^{1/4}=2e^{i 5\pi/8}$$
or
$$(-16)^{1/4}=2e^{i 9\pi/8}$$

is this correct?

As far as it goes. Where is your fourth 4th root?

the fourth root of -16 is $$2e^{i 9\pi/8}$$

didnt i show that?

Hurkyl
Staff Emeritus
Gold Member
If that's the fourth 4th root, then where's your third? You've only written three roots down, so you're missing at least one of them!

I'm not sure I understand. I am to find the 4th root of -16, not the fourth 4th root. and (-16)^(1/4) is the fourth root of -16, so I'm not sure what else is needed since $$(-16)^{1/4}=2e^{i 9\pi/8}$$

Hurkyl
Staff Emeritus
Gold Member
I am to find the 4th root of -16
In the original question, you said you're supposed to find the 4th roots of -16. As in all of them. How many 4th roots does -16 have? How many have you shown?

$$(-16)^{1/4}=2e^{i 9\pi/8}$$
I guess I'm not up on the convention for this stuff, but I would say that this is wrong. I would say the L.H.S. is multivalued, and denotes all fourth roots of -16, and the R.H.S. is a single value, denoting one fourth root of -16. Thus, it wouldn't be appropriate to write an equality there.

I see, so they are equal but just not equal in showing ALL the fourth roots of -16 right?

well that's like saying sin^-1(1/2) doesn't equal pi/6, because it also equals 5pi/6 and, well, so on

Hurkyl
Staff Emeritus
Gold Member
well that's like saying sin^-1(1/2) doesn't equal pi/6, because it also equals 5pi/6 and, well, so on
I agree. Lots of silly mistakes are made because people forget that the inverse of the sin function is multivalued.

HallsofIvy
Homework Helper
UrbanXrisis said:
I see, so they are equal but just not equal in showing ALL the fourth roots of -16 right?
What??? "they are all equal but just not equal"???

Any number has 4 distinct fourth (complex) roots. For example the fourth roots of 1 are 1, -1, i, and -i. You were asked to find all of the fourth roots of -16. ("i am to find the 4th roots of -16")
You only showed three in your original post.

$$(-16)^{1/4}=2i^{1/4}$$
is wrong. The principle root of 16 is, of course, 2 but -1 is not equal to i!
What you should have written was
$$(-16)^{1/4}= 2(-1)^{1/4}$$
Now, what are the 4 distinct fourth roots of -1?

$$(-1)^{1/4}=e^{i \pi /4}$$
$$(-1)^{1/4}=e^{9i \pi /4}$$
$$(-1)^{1/4}=e^{17i \pi /4}$$
$$(-1)^{1/4}=e^{25i \pi /4}$$

right? so that:

$$(-16)^{1/4}= 2e^{i \pi /4}$$
or
$$(-16)^{1/4}= 2e^{9i \pi /4}$$
or
$$(-16)^{1/4}= 2e^{17i \pi /4}$$
or
$$(-16)^{1/4}= 2e^{25i \pi /4}$$