Find the square root of i where i is the square root of (-1).
Show your working.
Thanks guys
Hello, anthmoo!
I assume you're not allowed to use DeMoivre's Theorem.
. . It would be too easy!
Let: .z .= .a + bi, where a and b are real numbers and z² = i.Find the square root of i where i = √(-1)
Then we have: .(a + bi)² .= .i
Expand: .a² + 2abi + b²i² .= .i
We have: .(a² - b²) + 2abi .= .i
Equate real and imaginary components:
. . a² - b² .= .0 .[1]
- - - .2ab . = .1 .[2]
[1] gives us: .b² = a² . → . b = ±a
Solve [2] for b: .b = 1/(2a)
Substitute into [1]: .a² - (1/2a)² .= .0 . → . a² - 1/(4a²) .= .0
Multiply by 4a²: .4a^4 - 1 .= .0 . → . a^4 .= .1/4
. . . . - . - . - . ._ . . . . . . . - . . ._
Hence: .a = ±√2/2 .and .b = ±√2/2
. . . . . . . . . _ . . . . . _
Therefore: .√i .= .±(√2/2)(1 + i)
I would like to interject. I may be mistaken, but the question is defined. (I wish I had my pre-calculus book with me to back this up) Here goes:
Let z = (a + bi)
Where (a + bi) = r(cosx + i*sinx)
It can be proven that z^n for any real value of n is:
z^n = (a + bi)^n = [r(cosx + i*sinx)]^n = r^n(cosx + i*sinx)^n = r^n(cos(nx) + i*sin(nx))
I don't remember the name of this theorem, but I'll try to modivate it:
Let n = 2
z^2 = (a+bi)^2 = [r(cosx + i*sinx)]^2
r^2(cosx + i*sinx)^2 = r^2[(cosx)^2 + 2i(sinx)cosx + i^2*(sinx)^2]
r^2{[(cosx)^2 - (sinx)^2] + i*(2sinx*cosx)}
Note: cos2x = (cosx)^2 - (sinx)^2 ... sin2x = 2(sinx)cosx
r^2(cos2x + i*sin2x)
Let n = 3
z^3 = z^2*z = r^2(cos2x + i*sin2x)*r(cosx + i*sinx) = r^3(cos2x*cosx + i*cosx*sin2x + i*cos2x*sinx + i^2*sin2x*sinx)
r^3[(cos2x*cosx - sin2x*sinx) + i*(cosx*sin2x + cos2x*sinx)]
Note: cos(2x+x) = cos2x*cosx - sin2x*sinx ... sin(2x+x) = sin2x*cosx + cos2x*sinx
r^3(cos3x + i*sin3x)
Let n = 4
...
r^4(cos4x + i*sin4x)
Let n = n
...
r^n(cos(nx) + i*sin(nx))
Let n = 1/2, a = 0, b = 1
z = (0 + i*1) = i
r = |z| = 1
tan(x) = 1/0 --> x = pi/2
z^n = (i)^n = r^n(cos(nx) + i*sin(nx))
z^(1/2) = (i)^(1/2) = r^(1/2)(cos(x/2) + i*sin(x/2)) = 1^(1/2)(cos(pi/4) + i*sin(pi/4)) = 1*(sqrt(2)/2 + i*sqrt(2)/2)
(i)^(1/2) = sqrt(2)/2 + i*sqrt(2)/2
Please do not argue with me I know what I am talking about. This type of math (about numbers) is something I understand very very well.
Next, do not listen to your precalculus book. It is not a pure book on math. It is a basic math book. It will not explain this detail strictly mathematically.
I recomment to search for a longgg thread about imaginary numbers over here somewhere.
Thus, if you want to think about taking a square root think of it as I do. As a solution to the equation x^2=i.
Wow. I'm sorry I've offended you. Please don't take anything I say offensively because it's not my intention to make you look wrong. I know you're smart. I have no doubt you know a lot about math. But I would like to point out that I'm not necessarily wrong because I don't know as much.
I can say that what I did was not incorrect. I don't disagree that the problem can be seen as x^2=i but I assure you there is a 'proven' theorem that allows for the roots of complex numbers to be taken. Whether these roots are defined I cannot argue, but they can be gotten by means other than x^n=i.
Square roots are only defined for the non-negative real numbers. The statement sqrt(-4)=2i is wrong as I explained a lot of times already. I do not care what the pre-calculus book says, it is a pre-calculus book, it is intended for high-school students.
One problem with square roots for the complex numbers is that they are not "well-defined". Basically, there are multiple values. Functions can only give a unique value, not multiple values.
For example, the Riemann surface of the complex function
is shown below. Note as you travel along the surface you get to a different point where you started at. That is a problem. Because it produces multiple values. De moiver's theorem is used to find roots as in zero's of the cyclic equation.
Note: The same type of problem occur with the logarithmic function on the complex numbers.