## MATH1901 Quizzes

Quiz 2: Polar form and roots of complex numbers
Question 1 Questions
What are the modulus and the principal argument of $-5-5i$? Exactly one option must be correct)
 a) $5$ and $\frac{5\pi }{4}$ b) $5$ and $-\frac{3\pi }{4}$ c) $5\sqrt{2}$ and $\frac{5\pi }{4}$ d) $5\sqrt{2}$ and $-\frac{3\pi }{4}$ e) $5\sqrt{2}$ and $\frac{3\pi }{4}$

Choice (a) is incorrect
Remember that if $z=x+iy$ then $|z|=\sqrt{{x}^{2}+{y}^{2}}$ and the principal argument of $z$ is greater than $-\pi$ and less than or equal to $\pi$.
Choice (b) is incorrect
Remember that if $z=x+iy$ then $|z|=\sqrt{{x}^{2}+{y}^{2}}$.
Choice (c) is incorrect
Remember that the principal argument of $z$ is greater than $-\pi$ and less than or equal to $\pi$.
Choice (d) is correct!
$|-5-5i|=\sqrt{{\left(-5\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{50}=5\sqrt{2}$.
Recall that the principal argument is greater than $-\pi$ and less than or equal to $\pi$.
Choice (e) is incorrect
Remember that the principal argument of $z$ is greater than $-\pi$ and less than or equal to $\pi$. Check which quadrant $-5-5i$ lies in, using a diagram.
Check all options corresponding to the polar form of $2-2i$. (Zero or more options can be correct)
 a) $2cis\left(-\frac{\pi }{4}\right)$ b) $2cis\left(\frac{7\pi }{4}\right)$ c) $2\sqrt{2}cis\left(-\frac{\pi }{4}\right)$ d) $2\sqrt{2}cis\left(\frac{\pi }{4}\right)$ e) $2\sqrt{2}cis\left(\frac{7\pi }{4}\right)$

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be True.
Correct!
1. False
2. False
3. True
4. False
5. True
The cartesian form of $8cis\left(\pi \right)$ is Exactly one option must be correct)
 a) 8 b) $8\pi$ c) $8-i$ d) $8i$ e) $-8$

Choice (a) is incorrect
Recall that if $z=rcis\theta$ then the cartesian form of $z$ is $rcos\theta +irsin\theta .$
Choice (b) is incorrect
Recall that if $z=rcis\theta$ then the cartesian form of $z$ is $rcos\theta +irsin\theta .$
Choice (c) is incorrect
Recall that if $z=rcis\theta$ then the cartesian form of $z$ is $rcos\theta +irsin\theta .$
Choice (d) is incorrect
Recall that if $z=rcis\theta$ then the cartesian form of $z$ is $rcos\theta +irsin\theta .$
Choice (e) is correct!
$8cis\pi =8cos\pi +i8sin\pi =-8+0i=-8$.
If $z=-1+i$ then $z$ expressed in polar form is Exactly one option must be correct)
 a) $\sqrt{2}cis\frac{\pi }{4}$ b) $\sqrt{2}cis\frac{3\pi }{4}$ c) $\frac{1}{\sqrt{2}}cis\left(-\frac{\pi }{4}\right)$ d) $cis\frac{\pi }{4}$ e) $cis\left(-\frac{\pi }{4}\right)$

Choice (a) is incorrect
Suppose $z=x+iy\ne 0$ has polar form $rcis\theta$. Then $r=\sqrt{{x}^{2}+{y}^{2}}$, $\frac{x}{r}=cos\theta$ and $\frac{y}{r}=sin\theta$.
Choice (b) is correct!
$\sqrt{2}\left(cos\frac{3\pi }{4}+isin\frac{3\pi }{4}\right)$
$=\sqrt{2}\left(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)=-1+i.$
Choice (c) is incorrect
Suppose $z=x+iy\ne 0$ has polar form $rcis\theta$. Then $r=\sqrt{{x}^{2}+{y}^{2}}$, $\frac{x}{r}=cos\theta$ and $\frac{y}{r}=sin\theta$.
Choice (d) is incorrect
Suppose $z=x+iy\ne 0$ has polar form $rcis\theta$. Then $r=\sqrt{{x}^{2}+{y}^{2}}$, $\frac{x}{r}=cos\theta$ and $\frac{y}{r}=sin\theta$.
Choice (e) is incorrect
Suppose $z=x+iy\ne 0$ has polar form $rcis\theta$. Then $r=\sqrt{{x}^{2}+{y}^{2}}$, $\frac{x}{r}=cos\theta$ and $\frac{y}{r}=sin\theta$.
An equivalent form of the complex number $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ is Exactly one option must be correct)
 a) $cos\frac{i\pi }{4}$ b) $\frac{1}{\sqrt{2}}cis\frac{i\pi }{4}$ c) $\frac{1}{\sqrt{2}}cis\frac{\pi }{4}$ d) $cis\frac{\pi }{4}$ e) None of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$cis\frac{\pi }{4}=cos\frac{\pi }{4}+isin\frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$.
Choice (e) is incorrect
If $z=-1+i$ and $w=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ then $zw$ equals Exactly one option must be correct)
 a) $-\sqrt{2}+\sqrt{2}i$ b) $-\sqrt{2}$ c) $0$ d) $\sqrt{2}i$ e) None of the above

Choice (a) is incorrect
Recall that if $z=a+ib$ and $w=c+id$ then $zw=ac-bd+i\left(ad+bc\right)$.
Choice (b) is correct!
Choice (c) is incorrect
Recall that if $z=a+ib$ and $w=c+id$ then $zw=ac-bd+i\left(ad+bc\right)$.
Choice (d) is incorrect
Recall that if $z=a+ib$ and $w=c+id$ then $zw=ac-bd+i\left(ad+bc\right)$.
Choice (e) is incorrect
Recall that if $z=a+ib$ and $w=c+id$ then $zw=ac-bd+i\left(ad+bc\right)$.
Suppose that $w=2cis\frac{\pi }{4}$. Check all options which equal ${w}^{4}$. (Zero or more options can be correct)
 a) $-16i$ b) $-16$ c) $2cis\pi$ d) $8cis\pi$ e) $16cis\pi$

There is at least one mistake.
For example, choice (a) should be False.
Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
There is at least one mistake.
For example, choice (b) should be True.
${w}^{4}={2}^{4}{\left(cos\frac{\pi }{4}+isin\frac{\pi }{4}\right)}^{4}=16\left(cos\pi +isin\pi \right)=-16.$
There is at least one mistake.
For example, choice (c) should be False.
Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
There is at least one mistake.
For example, choice (d) should be False.
Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
There is at least one mistake.
For example, choice (e) should be True.
${w}^{4}={2}^{4}{\left(cis\frac{\pi }{4}\right)}^{4}=16cis\pi .$
Correct!
1. False Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
2. True ${w}^{4}={2}^{4}{\left(cos\frac{\pi }{4}+isin\frac{\pi }{4}\right)}^{4}=16\left(cos\pi +isin\pi \right)=-16.$
3. False Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
4. False Recall that if $w=rcis\theta$ then ${w}^{4}={r}^{4}{\left(cis\theta \right)}^{4}={r}^{4}cis\left(4\theta \right).$
5. True ${w}^{4}={2}^{4}{\left(cis\frac{\pi }{4}\right)}^{4}=16cis\pi .$
It is known that the polynomial equation
${z}^{4}-4{z}^{3}+14{z}^{2}-36z+45=0$
has $3i$ and $2-i$ as two of its roots. What are the other two roots? Exactly one option must be correct)
 a) $3i,\phantom{\rule{1em}{0ex}}2+i$ b) $2-3i,\phantom{\rule{1em}{0ex}}i$ c) $-3i,\phantom{\rule{1em}{0ex}}2+i$ d) $1-3i,\phantom{\rule{1em}{0ex}}2+i$ e) There is not enough information to be able to work this out.

Choice (a) is incorrect
Recall that if $z=a+ib$ is a root of a polynomial with real coefficients then so is $\stackrel{̄}{z}=a-ib$.
Choice (b) is incorrect
Recall that if $z=a+ib$ is a root of a polynomial with real coefficients then so is $\stackrel{̄}{z}=a-ib$.
Choice (c) is correct!
Any polynomial equation with real coefficients has
non-real roots in complex conjugate pairs.
Choice (d) is incorrect
Recall that if $z=a+ib$ is a root of a polynomial with real coefficients then so is $\stackrel{̄}{z}=a-ib$.
Choice (e) is incorrect
Recall that if $z=a+ib$ is a root of a polynomial with real coefficients then so is $\stackrel{̄}{z}=a-ib$.
The 5th roots of $-1$ are Exactly one option must be correct)
 a) $cis\left(\frac{\pi }{5}\right)$, $cis\left(\frac{3\pi }{5}\right)$, $cis\left(-\frac{3\pi }{5}\right)$ and $cis\left(-\frac{\pi }{5}\right)$ b) $cis\left(±\frac{\pi }{5}\right)$, $cis\left(±\frac{3\pi }{5}\right)$, $-1$ c) $cis\left(±\frac{\pi }{10}\right)$, $cis\left(±\frac{3\pi }{10}\right)$ and $-1$ d) $-1$, $i$, $i+1$, $-i$ and $-i-1$ e) $cis\left(\frac{\pi }{5}\right)$, $cis\left(\frac{3\pi }{5}\right)$, $cis\left(\frac{7\pi }{5}\right)$ and $cis\left(\frac{9\pi }{5}\right)$

Choice (a) is incorrect
How many fifth roots of a number are there?
Choice (b) is correct!
Observe that the fifth roots of $-1$ are spaced at equal intervals of $2\frac{\pi }{5}$ around the unit circle.
Choice (c) is incorrect
Suppose $z=rcis\theta$, so that ${z}^{5}={r}^{5}cis\left(5\theta \right)$. Then ${r}^{5}cis\left(5\theta \right)=-1=1cis\left(\pi +2k\pi \right)$ where $k$ is any integer. Solve for $r$ and find all $\theta$ which satisfy this equation.
Choice (d) is incorrect
Suppose $z=rcis\theta$, so that ${z}^{5}={r}^{5}cis\left(5\theta \right)$. Then ${r}^{5}cis\left(5\theta \right)=-1=1cis\left(\pi +2k\pi \right)$ where $k$ is any integer. Solve for $r$ and find all $\theta$ which satisfy this equation.
Choice (e) is incorrect
How many fifth roots of a number are there?
Find the 8th roots of
$-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i.$
Exactly one option must be correct)
 a) $\phantom{\rule{1em}{0ex}}\frac{1}{\sqrt{2}}cis\left(±\frac{\pi }{4}\right)$, $\frac{1}{\sqrt{2}}cis\left(±\frac{3\pi }{4}\right)$, $±1$ and $±i$ b) $\phantom{\rule{1em}{0ex}}\sqrt{2}cis\left(±\frac{\pi }{4}\right)$, $\sqrt{2}cis\left(±\frac{3\pi }{4}\right)$, $±\sqrt{2}$ and $±i\sqrt{2}$ c) $\phantom{\rule{1em}{0ex}}cis\left(±\frac{\pi }{4}\right)$, $cis\left(±\frac{3\pi }{4}\right)$, $cis\left(±\frac{\pi }{2}\right)$ and $cis\left(±\pi \right)$ d) $cis\left(-\frac{29\pi }{32}\right)$, $cis\left(-\frac{21\pi }{32}\right)$, $cis\left(-\frac{13\pi }{32}\right)$, $cis\left(-\frac{5\pi }{32}\right)$, $cis\left(\frac{3\pi }{32}\right)$, $cis\left(\frac{11\pi }{32}\right)$, $cis\left(\frac{19\pi }{32}\right)$ and $cis\left(\frac{27\pi }{32}\right)$ e) $\phantom{\rule{1em}{0ex}}cis\left(±\frac{\pi }{4}\right)$, $cis\left(±\frac{3\pi }{4}\right)$, $cis\left(±\frac{5\pi }{4}\right)$ and $cis\left(±\frac{7\pi }{4}\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect