## MATH1001 Quizzes

Quiz 3: Polar exponential form and functions
Question 1 Questions
Chooose all options giving polar exponential form of the complex number $z=-1+\sqrt{3}i$. (Zero or more options can be correct)
 a) $2{e}^{-\pi i∕3}$ b) $2{e}^{\pi i∕3}$ c) $2{e}^{2\pi i∕3}$ d) $2{e}^{-4\pi i∕3}$ e) $2{e}^{4\pi i∕3}$

There is at least one mistake.
For example, choice (a) should be False.
Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
There is at least one mistake.
For example, choice (b) should be False.
Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
There is at least one mistake.
For example, choice (c) should be True.
The modulus of $z$ is $2$ and its principal argument is $2\pi ∕3$, ($z$ is in the second quadrant).
There is at least one mistake.
For example, choice (d) should be True.
The modulus of $z$ is $2$ and its principal argument is $2\pi ∕3$, ($z$ is in the second quadrant). However an argument is also $2\pi ∕3-2\pi =-4\pi i∕3$. This leads to the polar exponential form $2{e}^{-4\pi i∕3}$. There is no compulsion to use the principal argument when writing a complex number in polar form of any kind.
There is at least one mistake.
For example, choice (e) should be False.
Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
Correct!
1. False Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
2. False Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
3. True The modulus of $z$ is $2$ and its principal argument is $2\pi ∕3$, ($z$ is in the second quadrant).
4. True The modulus of $z$ is $2$ and its principal argument is $2\pi ∕3$, ($z$ is in the second quadrant). However an argument is also $2\pi ∕3-2\pi =-4\pi i∕3$. This leads to the polar exponential form $2{e}^{-4\pi i∕3}$. There is no compulsion to use the principal argument when writing a complex number in polar form of any kind.
5. False Plot $z$ in the complex plane to determine which quadrant it lies in. Its modulus is found using the formula $|a+ib|=\sqrt{{a}^{2}+{b}^{2}}$. Then find its argument $\theta$ using, for example, $tan\theta =\sqrt{3}∕\left(-1\right)$. However you must remember to adjust your answer to take account of the quadrant!
Find all solutions of the equation ${e}^{z}=-1+\sqrt{3}i$. Exactly one option must be correct)
 a) $z=ln2+2\pi i∕3$ b) $z={e}^{2}\left(cos\pi ∕3+isin\pi ∕3\right)$ c) $z={e}^{-1}\left(cos\left(2\pi ∕3\right)+isin\left(2\pi ∕3\right)\right)$ d) $\left\{z\in ℂ\mid z={e}^{2}+i\left(2\pi ∕3+2k\pi \right),k\in ℤ\right\}.$ e) None of the above

Choice (a) is incorrect
Don’t forget that when you equate polar forms of two equal complex numbers, you have to remember that arguments are determined only up to an integer multiple of $2\pi$.
Choice (b) is incorrect
Recall the definition: when $z=x+iy$ then ${e}^{z}={e}^{x}\left(cosy+isiny\right)={e}^{x}\phantom{\rule{0.3em}{0ex}}{e}^{iy}$.
Choice (c) is incorrect
Recall the definition: when $z=x+iy$ then ${e}^{z}={e}^{x}\left(cosy+isiny\right)={e}^{x}\phantom{\rule{0.3em}{0ex}}{e}^{iy}$.
Choice (d) is incorrect
Recall the definition: when $z=x+iy$ then ${e}^{z}={e}^{x}\left(cosy+isiny\right)={e}^{x}\phantom{\rule{0.3em}{0ex}}{e}^{iy}$.
Choice (e) is correct!
In fact, the set of solutions of ${e}^{z}=-1+\sqrt{3}i$ is the set $\left\{z\in ℂ\mid z=ln2+i\left(2\pi ∕3+2k\pi \right),k\in ℤ\right\}.$ Thus there are infinitely many solutions, each having the same real part and with imaginary parts spaced $2\pi$ apart.
The notation $f:A\to B$ means Exactly one option must be correct)
 a) the function $f$ takes all elements of set $A$ as inputs and produces all elements of set $B$ as outputs. b) the function $f$ takes all elements of set $A$ as inputs and the outputs will always be in set $B$. c) the function $f$ takes some subset of $A$ as inputs and the outputs will always be in set $B$. d) none of the above.

Choice (a) is incorrect
The range of $f$ is a subset of $B$ and is not necessarily equal to $B$.
Choice (b) is correct!
The notation $f:A\to B$ means that the function is well-defined on all elements of $A$ (its input values come from $A$). Its output values all lie in $B$. However $B$ need not be equal to the range. The range (the set of values which $f$ actually maps to) is a subset of $B$ and may be equal to $B$.
Choice (c) is incorrect
The set $A$ is equal to the domain of $f$.
Choice (d) is incorrect
What is the largest possible domain and the corresponding range of the following function?
$f\left(x\right)=ln\left(x+2\right)$
Exactly one option must be correct)
 a) Domain : $ℝ$, Range : $ℝ$ b) Domain : $\left(2,\infty \right)$, Range : $ℝ$ c) Domain : $\left(-2,\infty \right)$, Range : $\left[0,\infty \right)$ d) Domain : $\left[-2,\infty \right)$, Range : $\left[0,\infty \right)$ e) Domain : $\left(-2,\infty \right)$, Range : $ℝ$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
What is the largest possible domain and the corresponding range of the function
$f\left(x\right)=sin\left({e}^{x}\right)\phantom{\rule{1em}{0ex}}?$
Exactly one option must be correct)
 a) Domain : $\left(0,\infty \right)$, Range : $\left[-1,1\right]$ b) Domain : $ℝ$, Range : $\left[-1,1\right]$ c) Domain : $ℝ$, Range : $ℝ$ d) Domain : $\left(0,\infty \right)$, Range : $ℝ$ e) Domain : $\left[0,2\pi \right]$, Range : $\left[-1,1\right]$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
If $f\left(x\right)={x}^{2}$ and $g\left(x\right)=x+1$ then the composite function $\left(f\circ g\right)\left(x\right)$ is equal to Exactly one option must be correct)
 a) ${x}^{2}+1$ b) $\left({x}^{2}+1\right)+1$ c) ${\left({x}^{2}+1\right)}^{2}$ d) ${\left(x+1\right)}^{2}$ e) $\sqrt{x+1}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
If $f\left(x\right)=\sqrt{x}$, $g\left(x\right)=x+1$ and $h\left(x\right)={e}^{x}$ then $\left(\rightf\circ g\circ h\left)\right\left(x\right)$ is given by Exactly one option must be correct)
 a) $\sqrt{{e}^{x+1}}$ b) $\sqrt{{e}^{x}+1}$ c) ${e}^{\frac{x}{2}}+1$ d) ${e}^{\left(\sqrt{x}+1\right)}$ e) $\sqrt{{e}^{x}}+1$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
When $f\left(x\right)=\frac{-4+3x}{-3+2x}$, what is $f\left(f\left(x\right)\right)$? Exactly one option must be correct)
 a) $f\left(f\left(x\right)\right)=\frac{-24+x}{-3+2x}$ b) $f\left(f\left(x\right)\right)=\frac{x}{17}$ c) $f\left(f\left(x\right)\right)=\frac{3+24x}{16+2x}$ d) $f\left(f\left(x\right)\right)=4+6x$ e) $f\left(f\left(x\right)\right)=x$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
The formulas given below are formulas for various functions of the complex variable $z$. Which two functions have the same natural domain and corresponding range? $f\left(z\right)={e}^{z},\phantom{\rule{2em}{0ex}}g\left(z\right)={e}^{|z|},\phantom{\rule{2em}{0ex}}h\left(z\right)={e}^{{z}^{2}},\phantom{\rule{2em}{0ex}}k\left(z\right)={z}^{2}$ (Zero or more options can be correct)
 a) $f$ and $g$ b) $f$ and $k$ c) $f$ and $h$ d) $h$ and $k$ e) $g$ and $k$

There is at least one mistake.
For example, choice (a) should be False.
$f$ and $g$ have the same natural domain but different ranges.
There is at least one mistake.
For example, choice (b) should be False.
$f$ and $k$ have the same natural domain but different ranges.
There is at least one mistake.
For example, choice (c) should be True.
Both $f$ and $h$ have domain $ℂ$ and range $ℂ\setminus \left\{0\right\}$.
There is at least one mistake.
For example, choice (d) should be False.
$h$ and $k$ have the same natural domain but different ranges.
There is at least one mistake.
For example, choice (e) should be False.
$g$ and $k$ have the same natural domain but different ranges.
Correct!
1. False $f$ and $g$ have the same natural domain but different ranges.
2. False $f$ and $k$ have the same natural domain but different ranges.
3. True Both $f$ and $h$ have domain $ℂ$ and range $ℂ\setminus \left\{0\right\}$.
4. False $h$ and $k$ have the same natural domain but different ranges.
5. False $g$ and $k$ have the same natural domain but different ranges.
Find the natural domain and corresponding range of the function $f$ of the real variable $x$ given by the formula $f\left(x\right)=ln\left(x-{x}^{2}\right)$. Exactly one option must be correct)
 a) Natural domain $\left(0,1\right)$, range $\left(-\infty ,-ln4\right]$ b) Natural domain $\left(-1,1\right)$, range $\left(-\infty ,0\right]$ c) Natural domain $\left(0,1\right)$, range $\left(1,\infty \right)$ d) Natural domain $\left(-1,0\right)$, range $\left(-\infty ,ln1∕4\right]$ e) Natural domain $\left(0,1\right)$, range $\left(-\infty ,\frac{1}{4}\right)$

Choice (a) is correct!
We need $x-{x}^{2}>0$,and this occurs when $0, giving the domain $\left(0,1\right)$. The values of $x-{x}^{2}$ reach a maximum of $\frac{1}{4}$ when $x=\frac{1}{2}$, and as $x$ gets close to $0$ and to $1$, $x-{x}^{2}$ gets closer and closer to $0$. Therefore $f\left(x\right)$ has vertical asymptotes to $-\infty$ at $x=0$ and $x=1$, and has a maximum value of $ln\frac{1}{4}=-ln4$ when $x=\frac{1}{2}$. Hence the range is $\left(-\infty ,-ln4\right]$.
Choice (b) is incorrect
Remember that for the natural logarithm to be defined, it must be used with positive real inputs, so we need $x-{x}^{2}>0$.
Choice (c) is incorrect
The natural domain is correct, but the corresponding range is incorrect. Draw a rough sketch of the graph to help find the range.
Choice (d) is incorrect
Remember that for the natural logarithm to be defined, it must be used with positive real inputs, so we need $x-{x}^{2}>0$.
Choice (e) is incorrect
The natural domain is correct, but the corresponding range is incorrect. Draw a rough sketch of the graph to help find the range.