What is the square root of the square root of -16 in imaginary numbers?

Question

-16 as in negative 16...

Answer 1

Palindrome Geoff is right about how to arrive at the root he described, however, that is not the whole story.

Just as there are TWO numbers that satisfy the meaning of √16 (that is, 4 and -4) there are actually FOUR numbers that satisfy the meaning of √(√(-16)):
● √2 + i√2
● -√2 + i√2
● √2 - i√2
● -√2 - i√2

Geoff's analysis about "turning the angle halfway" toward the positive x axis is a good intuitive way of understanding what is going on. However, what he left out is that it DOESN'T MATTER which direction you turn, clockwise or counter-clockwise. Both directions are equally valid and both should be taken into account.

For √(-16), turning clockwise gives you 4i, counter-clockwise gives you -4i

For √(4i), turning clockwise gives you √2 + i√2, counter-clockwise gives you -√2 - i√2 (remember, you are going halfway to the positive x axis the long way)

For √(-4i), turning clockwise gives you -√2 + i√2, counter-clockwise gives you √2 - i√2

Answer 2

You no doubt know, a square root is a number that when multiplied by itself is equal to a given number. For example, 4 is the square root of 16, since 4 x 4 = 16. Note, however, that -4 x -4 = 16, too. We call 4 the positive square root of 16, and -4 the negative square root of 16. Now, you want to know if we can find the square root of a negative number. Let's take -16. We need to find a number, call it x, such that: x times x (x^2) = -16 Now, we know that any number times itself must be positive, not negative. Therefore, there is no such number x in the set of real numbers. A number x is defined, however, in the set of complex numbers. The complex numbers are a superset of the real numbers. That is, the complex numbers form a bigger set. The reals are a subset of the complex. A complex number has the form a + bi, where a and b are real numbers and the i is a special number. The "a" is called the real part; the "bi" is called the imaginary part. If we let a equal 0, then we have an imaginary number. The set of imaginary numbers is also a subset of the complex numbers. If we let b equal 0, then we have a regular real number. This is why the reals are a subset of the complex: the reals are just complex numbers that all have b=0, that is, no imaginary part. Now, the number i is defined to be equal to the square root of -1. This means that i^2 (i squared) is equal to -1. So now we can find the square root of -16. Since -16 = (-1) 16, we can write: sqr(-16) = sqr(-1) times sqr(16) (property of square roots) sqr(-16) = i times 4 This is usually written as 4i. We can check by squaring 4i. We get 4 x 4 = 16 times i x i = sqr(-1) times sqr(-1) = -1, giving 16 times -1 or -16. There is much more to the complex and imaginary sets of numbers than I can go into here. There are entire branches of mathematics (like complex analysis) which deal with these numbers so go buy you a book!!!! Gorge Cain comes to mind...

Answer 3

Oh. You said square root of the square root.

We know that the square root of -16 is 4i. So we take the square root of that again.

Use the general formula for square roots in imaginary space. Draw the real numbers on the x axis and the imaginary numbers on the y axis. To take a square root, divide your angle in half, and take the square root of the distance from the origin.

So -16 is 16 units away from the origin, at an angle of 180 degrees from the positive reals. So its square root is 4 at an angle of 90 degrees, which is 4i.

Then you take another square root. The distance from the origin is 4, so our new distance is 2. But we're at an angle of 45 degrees. If you plot this point and use the 45-45-90 triangle rule, you know that the hypotenuse (from the origin) is 2, so each side must be sqrt(2). So the answer is sqrt(2) + sqrt(2) i.

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The reason this trick works is because of the exponential form of complex numbers. Write your angle as T in radians, and the distance from the origin as R, and the complex number is R * e^(T i). We're practically defining the action of taking e to imaginary powers (there is a way to prove it, but it involves calculus and infinite series. Ask another question if you want me to demonstrate) so we say that this is true.

Now your original number was 16 * e^(pi i), because 180 degrees is pi radians. So take the square root: sqrt(16 * e^(pi i)) = 4 sqrt(e^(pi i)) = 4 (e^(pi i))^(1/2) = 4 e^(pi/2 i). So this new number is 4 angle pi/2, and pi/2 radians is 90 degrees.

Take the square root again: sqrt(4 e^(pi/2 i)) = 2 sqrt(e^(pi/2 i)) = 2 e^(pi/4 i). Now we just resolve this angle of pi/4 radians (45 degrees) back into the a+bi form, as above. So we get sqrt(2) + sqrt(2) i.

**Please note that answers like 4i and 2i are wrong. They don't consider that you're taking the square root of i also.**

Answer 4

The square roots of -16 are 4i and - 4i
The square root of those square roots are
√2(1+i), √2(1-i), √2(-1+i), √2(-1-i)

Answer 5

2i

The square root of 16 is 4, and the square root of 4 is 2. The square root of -1 does not exist, but is denoted mathmatically by the letter i.

Answer 6

-2

Answer 7

That would be 2 (imaginary)

Answer 8

RT11 guru is right just use the cardinal plane to find the phase and then you will find it even without a paper you just have to divide it 4 times

Answer 9

4i inches

Answer 10

4i