square root a complex number

[oberyn]

Hello everyone. Need help with the following;

Case A;

d(a+ib) = a+ib , i is a complex number
dt)

and

da = a and db = b
dt .............dt


Case B;

d(a+ib) =sqrt (a+ib)
(dt)

what is da and db in this case?
...........dt .......dt

any suggestion? Thanks.

 

[Deleted member 4993]

Hello everyone. Need help with the following;

Case A;

[U]d(a+ib)[/U] = a+[I]i[/I]b , [I]i[/I] is a complex number
   dt)

and 

[U]da[/U] = a  and [U]db[/U] = b  
dt              dt


Case B;

[U]d(a+[I]i[/I]b)[/U] =sqrt (a+[I]i[/I]b)
   (dt)

what is [U]da[/U]  and [U]db[/U] in this case?  
           dt        dt

any suggestion? Thanks.

How would you calculate df/dt, if f = √(x)

 

[oberyn]

How would you calculate df/dt, if f = √(x)

My intention is to solve the equation numerically.

if (x) is a function of time, the derivative of f = √(x) is straightforward & can be easily programmed.

in my case however, the i must be taken out from the sqrt so that it can be eliminated.
*note, a&b is a function of t.

 

[HallsofIvy]

I have no idea what you are doing or what question you are asking! a+ bi, I presume, is a general complex number (i itself is "the complex unit", not just "a complex number"). Although this is posted under "Calculus", surely that "d(a+ ib)/dt" cannot be the derivative? The derivative of a number (a constant) is 0. Are you asking, as Subhotosh Kahn suggests, for the derivative of \(\displaystyle f(z)= \sqrt{z}= z^{1/2}\)?

 

[oberyn]

I have no idea what you are doing or what question you are asking! a+ bi, I presume, is a general complex number (i itself is "the complex unit", not just "a complex number"). Although this is posted under "Calculus", surely that "d(a+ ib)/dt" cannot be the derivative? The derivative of a number (a constant) is 0. Are you asking, as Subhotosh Kahn suggests, for the derivative of \(\displaystyle f(z)= \sqrt{z}= z^{1/2}\)?

Ok. really sorry for the confusing question. let me try again,

lets say we have sample problem of the following form;

d(a+ib)/dt =a+b+ai+ 2bi

then,

da/dt = a+b
db/dt = a+2b

these equations are to be solve numerically, simultaneously.

I want to do the same for

d(a+ib)/dt =sqrt(a+ib)

the problem is, how do I eliminate i from the square root?

if it is sqrt(iab), i can be remove from the sqrt by using de moivre's theorem.

 

[HallsofIvy]

Are you saying that this "d/dt" has nothing to do with differentiation and that you are just assuming some function, f, such that f(a+ bi)= (a+ b)+ (a+ 2b)i? If so that is very confusing notation!

But your real question is about taking the square root of a complex number. The simplest way to take the square root of a complex number is to write the number if polar form. First write a+ bi as \(\displaystyle re^{i\theta}\) where \(\displaystyle r= \sqrt{a^2+ b^2}\) and \(\displaystyle \theta= arctan(b/a)\). Then \(\displaystyle \sqrt{a+ bi}= \sqrt{re^{i\theta}}= \sqrt{r}e^{i\theta/2}\).

 

[oberyn]

Are you saying that this "d/dt" has nothing to do with differentiation and that you are just assuming some function, f, such that f(a+ bi)= (a+ b)+ (a+ 2b)i? If so that is very confusing notation!

Not really. The d(a+bi)/dt actually appears as stochastic evolution used in quantum optics.

But your real question is about taking the square root of a complex number. The simplest way to take the square root of a complex number is to write the number if polar form. First write a+ bi as \(\displaystyle re^{i\theta}\) where \(\displaystyle r= \sqrt{a^2+ b^2}\) and \(\displaystyle \theta= arctan(b/a)\). Then \(\displaystyle \sqrt{a+ bi}= \sqrt{re^{i\theta}}= \sqrt{r}e^{i\theta/2}\).

Then,
sqrt (a+bi) = sqrt(r)cos (theta/2) + i sqrt(r)sin(theta/2),
can I use this form directly in the calculation?

 

[HallsofIvy]

Yes, that is correct. But, apparently this is a "differential equation" and you seem to be saying that you do not know anything about "differential equations"!

 

[oberyn]

Thanks anyway