This set of Complex Analysis Objective Questions & Answers focuses on “Differentiability”.
1. Which of the following is true?
a) Differentiability does not imply continuity
b) Differentiability implies continuity
c) Continuity implies differentiability
d) There is no relation between continuity and differentiable
View Answer
Explanation: Any function that is differentiable is definitely continuous. If a function is not continuous, it cannot be differentiable either. However, continuity cannot always imply differentiability. Sometimes, functions that are continuous are not differentiable.
2. Which of the following is correct about the function f(x)=|x+5|?
a) Left and right limits are equal and it is differentiable
b) Left and right limits are not equal and hence it is differentiable
c) Left and right limits are equal and it is not differentiable
d) Left and right limits are not equal and hence it is not differentiable
View Answer
Explanation: f(x)=(x+5), if x≥-5 and f(x)=-(x+5), if x<-5
Since the break is at x=-5, we calculate the limit at this point.
Left-hand limit: For x values slightly less than -5, (x+5) is negative. So, |x+5| will be -(x+5). By substituting x=-5, the limit will be -(-5+5) = 0.
Right-hand limit: For x values slightly more than -5, (x+5) is positive. So, |x+5| will be (x+5). By substituting x=-5, the limit will be (-5+5) = 0.
Hence, both left and right limits are same.
Derivative from the left: for x < -5, f(x) = -(x+5). So, the derivative value = -1.
Derivative from the right: for x ≥ -5, f(x) = (x+5). So, the derivative value = 1.
Since left and right derivatives are NOT-equal, the function is not differentiable.
3. Which of the following is correct about the function f(x)=|x2+18x+81|?
a) Left and right limits are equal and hence it is differentiable
b) Limits are not important to determine differentiability
c) Left and right limits are equal and hence it is not differentiable
d) Left and right limits are not equal and hence it is not differentiable
View Answer
Explanation: f(x)=|(x+9)2|
The expression inside the absolute value is always non-negative because (x+9) is squared. Squaring any number (positive or negative) results in a non-negative value (0 or a positive number). So, the absolute value function doesn’t introduce any sharp corners or discontinuities. As a result, the left and right-hand limits (which would normally be checked for differentiability) become irrelevant in this case. The function is differentiable due to absence of any sharp corners.
4. Which of the following is true about f(z)=z2?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable
d) Differentiable but not continuous
View Answer
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x,y)=(x+iy)2
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)(x+iy)2
\(\lim_{y \to 0}\)i2y2
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)(x+iy)2
\(\lim_{x \to 0}\)x2
=0
Both the limits are equal, therefore the function is continuous. To check differentiability,
f’(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\)should exist.
\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2-(z)^2}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{2z(\delta z)+(\delta z)^2}{\delta z}\)
=\(\lim_{\delta z \to 0}\)(2z+(δz))
=2z
Since f’(z) exists, the function is differentiable as well.
5. Which of the following is true about f(z)=z+iz?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable
d) Differentiable but not continuous
View Answer
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x, y)=x+iy+ix+i2y
f(x, y)=(x-y)+i(x+y)
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)(x-y)+i(x+y)
\(\lim_{y \to 0}\)-y+iy
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)x+iy+ix+i2y
\(\lim_{x \to 0}\)x+ix
= 0
Both the limits are equal, therefore the function is continuous. To check differentiability,
f’(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\) should exist.
\(\lim_{\delta z \to 0}\frac{z+\delta z+iz+i\delta z-z+iz}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{\delta z+i\delta z}{\delta z}\)
=z+i
Since f’(z) exists, the function is differentiable as well.
6. Which of the following is true about f(z)=z2+2z?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable
d) Differentiable but not continuous
View Answer
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x, y)=(x+iy)2+2(x+iy)
f(x, y)=x2-y2+2xiy+2x+2iy
Left limit: \(\lim_{x \to 0 \\ y \to 0}\)x2-y2+2xiy+2x+2iy
\(\lim_{y \to 0}\)-y2+2iy
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\)x2-y2+2xiy+2x+2iy
\(\lim_{x \to 0}\)x2+2x
=0
Both the limits are equal, therefore the function is continuous. To check differentiability,
f’(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\) should exist.
\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2+2(z+\delta z)-(z^2+2z)}{\delta z}\)
\(\lim_{\delta z \to 0}\frac{z^2+(\delta z)^2+2z(\delta z)+2z+2(\delta z)-z^2-2z}{\delta z}\)
=2z+2
Since f’(z) exists, the function is differentiable as well.
7. Which of the following is true about f(z)=\(\frac{z+iz}{z^2}\)?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable
d) Differentiable but not continuous
View Answer
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x, y)=\(\frac{x+iy+ix+i^2y}{(x+iy)^2}\)
f(x, y)=\(\frac{i(x+y)+(x-y)}{(x+iy)^2}\)
Left limit: \(\lim_{x \to 0 \\ y \to 0}\frac{i(x+y)+(x-y)}{(x+iy)^2}\)
\(\lim_{y \to 0}\frac{i(y)+(-y)}{(iy)^2}\)
=does not exist
Since, the left limit itself does not exist, the function is not continuous. If a function is not continuous, it cannot be differentiable as well.
8. Which of the following is true about f(z)=\(\frac{z^2+(iz)^2}{z^2}\)?
a) Continuous and differentiable
b) Continuous but not differentiable
c) Neither continuous nor differentiable
d) Differentiable but not continuous
View Answer
Explanation: z=x+iy
In general the limits are discussed at origin, if nothing is specified.
f(x, y)=\(\frac{(x+iy)^2+(ix+i^2y)^2}{(x+iy)^2}\)
f(x, y)=\(\frac{(x+iy)^2+(ix-y)^2}{(x+iy)^2}\)
Left limit: \(\lim_{x \to 0 \\ y \to 0}\frac{(x+iy)^2+(ix-y)^2}{(x+iy)^2}\)
\(\lim_{y \to 0}\frac{(iy)^2+(-y)^2}{(iy)^2}\)
\(\lim_{y \to 0}\frac{-(y)^2+(y)^2}{-(y)^2}\)
=\(\frac{-1+1}{-1}\)
=0
Right limit: \(\lim_{y \to 0 \\ x \to 0}\frac{(x+iy)^2+(ix-y)^2}{(x+iy)^2}\)
\(\lim_{x \to 0}\frac{(x)^2+(ix)^2}{(x)^2}\)
\(\lim_{y \to 0}\frac{(x)^2-(x)^2}{(x)^2}\)
=\(\frac{1-1}{1}\)
=0
Both the limits are equal, therefore the function is continuous. To check differentiability,
f(z)=\(\frac{z^2-z^2}{z^2}\)=0
f’(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\)should exist.
\(\lim_{\delta z \to 0}\frac{0-0}{\delta z}\)
=0
Since f’(z) exists, the function is differentiable as well.
Sanfoundry Global Education & Learning Series – Complex Analysis.
To practice all objective questions on Complex Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.
- Check Engineering Mathematics Books
- Practice Numerical Methods MCQ
- Apply for 1st Year Engineering Internship
- Practice Probability and Statistics MCQ
