What are the 4th roots of -16?

[UrbanXrisis]

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?

 

[TMFKAN64]

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

 

[UrbanXrisis]

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

didnt i show that?

 

[Hurkyl]

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!

 

[UrbanXrisis]

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]

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.

 

[UrbanXrisis]

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

 

[schattenjaeger]

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]

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]

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.

Actually, your very first statement:
(-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?

 

[UrbanXrisis]

(-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}