Complex Function Theory Questions and Answers – Differentiability

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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

Answer: b
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.
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2. Which of the following is correct about the function f(x)=|x+5|?
a) Left and right limits are equal and hence it is differentiable
b) Left and right limits are not equal and hence it is differentiable
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

Answer: d
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.
We know that, a function is not differentiable at point x=a, if either \(\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}\)does not exist or is infinity. We check limits for both the cases of the function.
Left limit: Here, a=-5.
\(\lim_{h \to 0}\frac{f(-5+h)-f(-5)}{h}\)
\(\lim_{h \to 0}\frac{-(-5+h+5)-(-(-5+5))}{h}\)
=-1
Right limit: Here, a=1.
\(\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}\)
\(\lim_{h \to 0}\frac{(1+h+5)-(1+5)}{h}\)
=1
Since the two limits 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

Answer: d
Explanation: f(x)=|(x+9)2|
Since the break is at x=-9, we frame for the function for both the sides of -9.
f(x)=(x+9)2, if x>-9 and f(x)=(x+9)2, if x≤-9
We know that, a function is not differentiable at point x=a, if either \(\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}\)does not exist or is infinity. We check limits for both the cases of the function.
Left limit: Here, a=-10.
\(\lim_{h \to 0}\frac{f(-10+h)-f(-10)}{h}\)
\(\lim_{h \to 0}\frac{(-10+h+9)^2-(-10+9)^2}{h}\)
\(\lim_{h \to 0}\frac{1^2+h^2-2h-1^2}{h}\)
=-2
Right limit: Here, a=10.
\(\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}\)
\(\lim_{h \to 0}\frac{(10+h+9)^2-(10+9)^2}{h}\)
\(\lim_{h \to 0}\frac{10^2+h^2+20h-10^2}{h}\)
=20
Since the two limits are not equal, the function is not differentiable.
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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

Answer: a
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

Answer: a
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.
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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

Answer: a
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

Answer: c
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.
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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

Answer: a
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.

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