Implicit Differentiation and Complex numbers concept question

[DrummingAtom]

I just started learning Implicit Differentiation and came across an issue. I took the derivative of the circle function:

y² + x² = 1

y' = -x / y

This all made sense until I solved the circle function for y, which gives:

y = \pm\sqrt{1 - x^2}

For any x > 1, it's going to be complex. So, does this mean that for any x > 1 I can now compute the derivative for those complex numbers? Because plugging a x > 1 into my derivative formula I'll get a number. Any help is appreciated, thanks.

 

[tiny-tim]

Hi DrummingAtom!

Let's use different letters, to make it easier …

if z² + t² = 1, with t real and z complex,

then dz/dt = -t/z, and everything's ok

(try it with z = x + iy if you're not happy)

 

[Wzet]

I don't know if I'm right... But as I see it, circle is not defined for any x > 1 since radius of this circle is 1. At a point x > 1 there is no circle so I don't think it's of any use..

 

[tiny-tim]

Welcome to PF!

Hi Wzet! Welcome to PF!

Yes, the curve doesn't go into the x > 1 region, if we allow an ordinary (real) y-axis.

But if we allow a "complex y-axis", the curve y = √(1 - x²) carries on up to "infinite" x.

 

[Wzet]

Thanks for welcoming :)

I thought the question was, if it helpful and if the computation of derivative when x > 1 tells us something. if x < 1 then we can compute useful things, but is the complex derivative of the function helpful?

 

[tiny-tim]

… is the complex derivative of the function helpful?

sometimes

(one of the advantages of complex functions is that they do most of the things that real functions do … only more so! )

 

[HallsofIvy]

I just started learning Implicit Differentiation and came across an issue. I took the derivative of the circle function:

y² + x² = 1

y' = -x / y

This all made sense until I solved the circle function for y, which gives:

y = \pm\sqrt{1 - x^2}

For any x > 1, it's going to be complex. So, does this mean that for any x > 1 I can now compute the derivative for those complex numbers? Because plugging a x > 1 into my derivative formula I'll get a number. Any help is appreciated, thanks.

If x> 1, then (x, y) is not on the circle so your equation is invalid.

 

[DrummingAtom]

I need to clarify my question more. I'm new to complex numbers and have only done very basic things with them, so bear with me.

After plotting some complex number points of the function t =<br /> \sqrt{1 - z^2}<br />

It appears to have positive slope, but according to the dz/dt = -t/z, it should be negative. How is this working? Thanks.

 

[tiny-tim]

Hi DrummingAtom!

I'm not sure what you've drawn,

but if t is real, and if z is "positive" imaginary (z = iy), then 1/z is "negative" imaginary (z = -i/y), and -t/z is "positive" imaginary.

Does that help?​

 

[jackmell]

I just started learning Implicit Differentiation and came across an issue. I took the derivative of the circle function:

y² + x² = 1

y' = -x / y

This all made sense until I solved the circle function for y, which gives:

y = \pm\sqrt{1 - x^2}

For any x > 1, it's going to be complex. So, does this mean that for any x > 1 I can now compute the derivative for those complex numbers? Because plugging a x > 1 into my derivative formula I'll get a number. Any help is appreciated, thanks.

Yes, except the derivative does not exist at \pm 1. The circle is only a small part of the more global complex function. To see that, write it as:

w^2+z^2=1

and treat w and z as complex numbers. We can solve for w:

w=\sqrt{1-z^2}

where it is now understood the square root is multi-valued and now represents an object in complex space. The "real" circle we represent as x^2+y^2=1[/tex] is only the set of points on this &quot;complex&quot; object when w and z are real and satisfy the expession. <br /> <br /> Now, that complex object has a (multi) derivative everywhere except at z=\pm 1<br /> <br /> <br /> For example:<br /> <br /> \frac{dw}{dz}\biggr|_{z=-2}=-\frac{z}{\sqrt{1-z^2}}\biggr|_{z=-2}=\frac{2}{\sqrt{-3}}=\frac{2}{\pm i \sqrt{3}}=\pm i\frac{2}{\sqrt{3}}.<br /> <br /> That is now a &quot;complex&quot; derivative and since the function was multivalued, so too is it&#039;s derivative. However, this derivative of a complex function no longer represents a &quot;slope&quot; of a curve but rather something more applicable to complex numbers.<br /> <br /> Here&#039;s your circle dude: See it right? In there. That figure is the real part of the &quot;complex object&quot; I was referring to.