Is f(x) = (x-iy)/(x-1) a Continuous Function?

In summary: Formally, you should see a theorem in your book that for f(x) and g(x) continuous, f(x)/g(x) is continuous except where g(x) = 0. The limit of f(x)/g(x) may exist as g(x)->0, for instance in the function sin(x) / x, but the function itself is not continuous there.
  • #1
Krayfish
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Homework Statement


Determine if the following function is continuous: f(x) = (x-iy)/(x-1)

Homework Equations


How do find out if a function is continuous without graphing it and without a point to examine? I know I've learned this, probably in pre-calculus too, but I'm blanking

The Attempt at a Solution


u(x) = x-iy as a function is continuous because, due to the i term, x-iy will never equal 0 and it is a linear function
v(x) = x-1 is also continuous,as it is a linear function that exists under all conditions

if u and v are both continuous under all conditions, than u/v must also be continuous?
 
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  • #2
From the point of view of precalculus, you have learned about limits, in particular, one-sided limits. What happens at x = 1?
 
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  • #3
Krayfish said:
if u and v are both continuous under all conditions, than u/v must also be continuous?
u/v will be continuous at points for which v ≠ 0...
 
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  • #4
Informally, you can't divide by 0. The function may blow up to +-infinity there. At any rate, it's not defined when the denominator is 0.

Formally, you should see a theorem in your book that for f(x) and g(x) continuous, f(x)/g(x) is continuous except where g(x) = 0. The limit of f(x)/g(x) may exist as g(x)->0, for instance in the function sin(x) / x, but the function itself is not continuous there. (This is a fine point called a "removable discontinuity", which is also in your textbook. Basically, if the limit exists, then you define the function to say "and when the denominator = 0, it's defined to be the limit")
 
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  • #5
Krayfish said:

Homework Statement


Determine if the following function is continuous: f(x) = (x-iy)/(x-1)
Did you mean ##f(z)##? If not, then how is ##y## defined?

u(x) = x-iy as a function is continuous because, due to the i term, x-iy will never equal 0 and it is a linear function
Does it matter if ##u## is ever equal to 0?
 
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  • #6
verty said:
From the point of view of precalculus, you have learned about limits, in particular, one-sided limits. What happens at x = 1?
OH yeah that one was pretty obvious, that's what you get when you're doing math with no sleep. Undefined, thank you
 

1. What is continuity in complex variables?

Continuity in complex variables refers to the smoothness and connectedness of a complex function. It means that as the input values of the function change, the output values change in a predictable and continuous manner.

2. How is continuity different in complex variables compared to real variables?

In complex variables, continuity is more complex and requires the use of the Cauchy-Riemann equations to determine if a function is continuous. In real variables, continuity is more straightforward and can be determined using the limit definition.

3. What is the Cauchy-Riemann condition?

The Cauchy-Riemann condition is a set of necessary and sufficient conditions for a complex function to be differentiable. It states that the partial derivatives of the real and imaginary parts of the function must exist and satisfy certain relationships at a given point in order for the function to be differentiable at that point.

4. Can a function be continuous but not differentiable in complex variables?

Yes, it is possible for a function to be continuous but not differentiable in complex variables. This can occur at points where the Cauchy-Riemann condition is not satisfied or when the function has a singularity.

5. How is the concept of continuity used in complex analysis?

In complex analysis, continuity is an important concept as it is used to determine the properties of complex functions, such as differentiability and analyticity. It also helps in understanding the behavior of complex functions and their mappings.

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