nyuuzyou/wikis huggingface dataset:
Root of Number Greater than One

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \N_{>0}$ be a natural number such that $n > 0$.
Then:
:$x \ge 1 \implies x^{1/n} \ge 1$
where $x^{1/n}$ is the $n$th root of $x$.
== Proof ==
Let $y = x^{1/n}$.
From the definition of the $n$th root of $x$, it follows that $x = y^n$.
We will show by induction that $\forall n \in \N_{>0}: y^n \ge 1 \implies y \ge 1$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$y^n \ge 1 \implies y \ge 1$
=== Basis for the Induction ===
By definition, $y^1 = y$.
Thus $\map P 1$ is true, as this just says $y \ge 1 \implies y \ge 1$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So our induction hypothesis is that:
:$y^k \ge 1 \implies y \ge 1$
Now we need to show that:
:$y^{k + 1} \ge 1 \implies y \ge 1$
=== Induction Step ===
This is our induction step:
By definition:
:$y^{k + 1} = y \cdot y^k$
Suppose $y^k \ge 1$.
From the induction hypothesis it follows that $y > 1$.
As $y \ge 1$ it follows that $y > 0$.
Let:
:$y^{k + 1} = y \cdot y^k \ge 1$
Then:
:$y \cdot y^k$
By Real Number Ordering is Compatible with Multiplication:
:$y \cdot y^k \ge y \times 1$
and hence the result.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: y^n \ge 1 \implies y \ge 1$
As $y^n = x$ and $y = x^{1/n}$, the result follows.
Category:Analysis
Category:Inequalities
Root of Quotient equals Quotient of Roots

== Theorem ==
Let $a, b \in \R_{>0}$ be (strictly) positive real numbers.
Let $n \in \Z_{>0}$ be a strictly positive integer
Then:
:$\sqrt [n] {\dfrac a b} = \dfrac {\sqrt [n] a} {\sqrt [n] b}$
where $\sqrt [n] {}$ denotes the $n$th root.
== Proof ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.9$
* : $\S 13$: Laws of Exponents: $13.9.$
Category:Exponent Combination Laws
Root of Reciprocal is Reciprocal of Root

== Theorem ==
Let $x \in \R_{\ge 0}$.
Let $n \in \N$.
Let $\sqrt [n] x$ denote the $n$th root of $x$.
Then:
:$\sqrt [n] {\dfrac 1 x} = \dfrac 1 {\sqrt [n] x}$
== Proof ==
Let $y = \sqrt [n] {\dfrac 1 x}$.
Then:
Hence the result.
Category:Roots of Numbers
Rooted Tree Corresponds to Arborescence

== Theorem ==
Let $T = \struct {V, E}$ be a rooted tree with root $r$.
Then there is a unique orientation of $T$ which is an $r$-arborescence.
== Proof ==
Recall that a tree is connected and has no cycles.
Thus there is exactly one path from each node of $T$ to each other node of $T$.
Let $A$ be the set of all ordered pairs $x, y \in V$ such that:
:$\tuple {x, y} \in E$ and
:The unique path from $r$ to $y$ passes through $x$.
== Sources ==
* : $\S 2.3.4.2$
Category:Rooted Trees
Roots of Complex Number

== Theorem ==
Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then the $n$th roots of $z$ are given by:
:$z^{1 / n} = \set {r^{1 / n} \paren {\map \cos {\dfrac {\theta + 2 \pi k} n} + i \, \map \sin {\dfrac {\theta + 2 \pi k} n} }: k \in \set {0, 1, 2, \ldots, n - 1} }$
There are $n$ distinct such $n$th roots.
=== Exponential Form ===
This result can also be expressed thus in exponential form:
=== Corollary ===
== Proof ==
Let:
: $w := r^{1 / n} \paren {\map \cos {\dfrac {\theta + 2 \pi k} n} + i \, \map \sin {\dfrac {\theta + 2 \pi k} n} }$
for $k \in \Z_{>0}$.
Then:
Now let $m = k + n$.
Then:
exploiting the fact that Sine and Cosine are Periodic on Reals.
The result follows.
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 3$. Geometric Representation of Complex Numbers
* : $\S 6$: Complex Numbers: Roots of Complex Numbers: $6.10$
* : $1$: Complex Numbers: Roots of Complex Number
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $80 \ \text{(a)}$
* : $\S 4$: Complex Numbers: Roots of Complex Numbers: $4.11.$
Category:Complex Analysis
Category:Complex Roots
Roots of Complex Number/Corollary

== Theorem ==
Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $w$ be one of the complex $n$th roots of $z$.
Then the $n$th roots of $z$ are given by:
:$z^{1 / n} = \set {w \epsilon^k: k \in \set {1, 2, \ldots, n - 1} }$
where $\epsilon$ is a primitive $n$th root of unity.
== Proof ==
By definition of primitive complex $n$th root of unity:
:$\omega = e^{2 m i \pi k}$
for some $m \in \Z: 1 \le m < n$.
Thus:
This demonstrates that $w \omega^k$ is one of the complex $n$th roots of $z$.
All of the complex $n$th roots of unity are represented by powers of $\omega$.
Thus it follows from Roots of Complex Number that:
:$z^{1 / n} = \set {w \omega^k: k \in \set {1, 2, \ldots, n - 1} }$
are the $n$ complex $n$th roots of $z$.
== Examples ==
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots
Roots of Complex Number/Corollary/Examples/Cube Roots

== Example of Roots of Complex Number: Corollary ==
Let $z \in \C$ be a complex number.
Let $z \ne 0$.
Let $w$ be one of the (complex) cube roots of $z$.
Then the complete set of (complex) cube roots of $z$ is:
:$\set {w, w \omega, w \omega^2}$
where:
:$\omega = e^{2 \pi / 3} = -\dfrac 1 2 + \dfrac {i \sqrt 3} 2$
== Proof ==
From Primitive Complex Cube Roots of Unity, $\omega$ is a primitive root of $1$.
Hence the result from Roots of Complex Number: Corollary.
== Sources ==
* : Introduction
Category:Examples of Complex Roots
Roots of Complex Number/Corollary/Examples/Cube Roots of 8i

== Example of Roots of Complex Number: Corollary ==
The complex $4$th roots of $8 i$ are given by:
:$\paren {2 - 2 i}^{1/4} = \set {2 i, \sqrt 3 + i, -\sqrt 3 + i}$
== Proof ==
Let $z = 8 i$.
Then:
:$z = 8 \exp \paren {\dfrac {i \pi} 2}$
Let $b$ be defined as:
Then we have that the complex cube roots of unity are:
:$1, \exp {\dfrac {2 i \pi} 3}, \exp {\dfrac {-2 i \pi} 3}$
Thus from Roots of Complex Number: Corollary:
== Sources ==
* : $\S 3$. Roots of Unity: Example $4$.
Category:Examples of Complex Roots
Roots of Complex Number/Corollary/Examples/Fourth Roots of 2-2i

== Example of Roots of Complex Number: Corollary ==
The complex $4$th roots of $2 - 2 i$ are given by:
:$\paren {2 - 2 i}^{1/4} = \set {b, bi, -b, -bi}$
where:
:$b = \sqrt [8] 8 \paren {\cos \dfrac \pi {16} + i \sin \dfrac \pi {16} }$
== Proof ==
Let $z = 2 - 2 i$.
Then:
and so:
:$\arg z = -\dfrac \pi 4$
Let $b$ be defined as:
Then we have that the complex $4$th roots of unity are:
:$1, i, -1, -i$
The result follows from Roots of Complex Number: Corollary.
== Sources ==
* : $\S 3$. Roots of Unity: Example $3$.
Category:Examples of Complex Roots
Roots of Complex Number/Examples

== Examples of Roots of Complex Number==
Note the use of the notation $\cis \theta$ in the below, which means $\cos \theta + i \sin \theta$.
=== Square Roots of $2 \sqrt 3 - 2 i$ ===
=== Square Roots of $4 \sqrt 2 + 4 \sqrt 2 i$ ===
=== Cube Roots of Complex Number ===
=== Cube Roots of $8$ ===
=== Cube Roots of $2 + 2 i$ ===
=== Cube Roots of $-1 + i$ ===
=== Cube Roots of $-11 - 2 i$ ===
=== Cube Roots of $2 + 11 i$ ===
=== Cube Roots of $2 - 11 i$ ===
=== Cube Roots of $2 + 2 \sqrt 3 i$ ===
=== Square of Cube Roots of $i$ ===
=== $4$th Roots of $i$ ===
=== $4$th Roots of $-16 i$ ===
=== $4$th Roots of $-2 \sqrt 3 - 2 i$ ===
=== Roots of $z^4 + 81 = 0$ ===
=== Roots of $z^4 - 81 = 0$ ===
=== $5$th Roots of $-32$ ===
=== $5$th Roots of $-4 + 4 i$ ===
=== $5$th Roots of $-16 + 16 \sqrt 3 i$ ===
=== Roots of $z^5 + 1 = 0$ ===
=== $6$th Roots of $64$ ===
=== $6$th Roots of $-27 i$ ===
=== Roots of $z^6 + 1 = \sqrt 3 i$ ===
=== Roots of $z^8 + 1 = 0$ ===
Category:Examples of Complex Roots
Roots of Complex Number/Examples/4th Roots of -16 i

== Example of Roots of Complex Number: Corollary ==
The complex $4$th roots of $-16 i$ are given by:
:$\paren {-16 i}^{1/4} = \set {2 \, \map \cis {67.5 + 90 k} \degrees}$
for $k = 0, 1, 2, 3$.
That is:
== Proof ==
:
Let $z^4 = -16 i$.
We have that:
:$z^4 = 16 \, \map \cis {\dfrac {3 \pi} 2 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(d)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/4th Roots of -2 root 3 - 2i

== Example of Roots of Complex Number: Corollary ==
The complex $4$th roots of $-2 \sqrt 3 - 2 i$ are given by:
:$\paren {-2 \sqrt 3 - 2 i}^{1/4} = \set {\sqrt 2 \paren {\map \cis {\dfrac {7 \pi} {24} + \dfrac {k \pi} 2} } }$
for $k = 0, 1, 2, 3$.
That is:
== Proof ==
:
Let $z^4 = -2 \sqrt 3 - 2 i$.
We have that:
:$z^4 = 4 \paren {\map \cis {\dfrac {7 \pi} 6 + 2 k \pi} }$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $29 \ \text {(b)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/4th Roots of i

== Example of Roots of Complex Number: Corollary ==
The complex $4$th roots of $i$ are given by:
:$i^{1/4} = \set {b, bi, -b, -bi}$
where:
:$b = \paren {\cos \dfrac \pi 8 + i \sin \dfrac \pi 8}$
== Proof ==
Let $z = i$.
Then:
and so:
:$\arg z = \frac \pi 2$
Let $b$ be defined as:
Then we have that the complex $4$th roots of unity are:
:$1, i, -1, -i$
The result follows from Roots of Complex Number: Corollary.
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $2$.
Category:Examples of Complex Roots
Roots of Complex Number/Examples/5th Roots of -16 + 16 root 3 i

== Example of Roots of Complex Number ==
The complex $5$th roots of $-16 + 16 \sqrt 3 i$ are given by:
:$\paren {-16 + 16 \sqrt 3 i}^{1/5} = \set {2 \, \map \cis {24 + 72 k} \degrees}$
for $k = 0, 1, 2, 3, 4$.
That is:
== Proof ==
:
Let $z^5 = -16 + 16 \sqrt 3 i$.
We have that:
:$z^5 = 16 \sqrt 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi} = 16 \sqrt 2 \, \map \cis {120 \degrees + k \times 360 \degrees}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(c)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/5th Roots of -16 + 16 root 3 i/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(c)}$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:Find all the indicated roots and locate them in the complex plane.
::... $\text {(c)}$ fifth roots of $-16 + 16 \sqrt 3 i$, ...
:Ans. ... $\text {(c)}$ $2 \cis 48 \degrees$, $2 \cis 120 \degrees$, $2 \cis 192 \degrees$, $2 \cis 264 \degrees$, $2 \cis 336 \degrees$ ...
== Correction ==
The correct solution is:
:$2 \cis 24 \degrees$, $2 \cis 96 \degrees$, $2 \cis 168 \degrees$, $2 \cis 240 \degrees$, $2 \cis 312 \degrees$
as demonstrated in $5$th Roots of $-16 + 16 \sqrt 3 i$.
The mistake probably arose from mistaking $\dfrac {2 \pi} 3 = 240 \degrees$ when calculating the arguments in degrees.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(c)}$
Category:Mistakes/Theory and Problems of Complex Variables
Roots of Complex Number/Examples/5th Roots of -32

== Example of Roots of Complex Number ==
Let $z^5 = -32$.
Then:
:$z = 2 \paren {\map \cos {\dfrac {\pi + 2 k \pi} 5} + i \, \map \sin {\dfrac {\pi + 2 k \pi} 5} }$
for $k = 0, 1, 2, 3, 4$.
That is:
== Proof ==
:
In polar form:
:$-32 = \polar {32, \pi + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $28$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/5th Roots of -4 + 4i

== Example of Roots of Complex Number ==
The complex $5$th roots of $-4 + 4i$ are given by:
:$\paren {-4 + 4i}^{1/5} = \set {\sqrt 2 \, \map \cis {27 + 72 k} \degrees}$
for $k = 0, 1, 2, 3, 4$.
That is:
== Proof ==
:
Let $z^5 = -4 + 4 i$.
We have that:
:$z^5 = 4 \sqrt 2 \, \map \cis {\dfrac {3 \pi} 4 + 2 k \pi} = 4 \sqrt 2 \, \map \cis {135 \degrees + k \times 360 \degrees}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(b)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/6th Roots of -27 i

== Example of Roots of Complex Number ==
The complex $6$th roots of $-27 i$ are given by:
:$\paren {-27 i}^{1/6} = \set {\sqrt 3 \, \map \cis {45 + 60 k} \degrees}$
for $k = 0, 1, 2, 3, 4, 5$.
That is:
== Proof ==
:
Let $z^6 = -27 i$.
We have that:
:$z^6 = 27 \, \map \cis {\dfrac {3 \pi} 2 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(d)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/6th Roots of 64

== Example of Roots of Complex Number ==
The complex $6$th roots of $64$ are given by:
:$\paren {64}^{1/6} = \set {2 \, \map \cis {60 k} \degrees}$
for $k = 0, 1, 2, 3, 4, 5$.
That is:
== Proof ==
:
Let $z^6 = 64$.
We have that:
:$z^6 = 64 \, \map \cis {0 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
When $k = 3$ we have:
:$z_4 = 2 \cis 180 \degrees = - 2$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(e)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots

== Theorem ==
Let $z := \polar {r, \theta}$ be a complex number expressed in polar form, such that $z \ne 0$.
Then the complex cube roots of $z$ are given by:
:$z^{1 / 3} = \set {r^{1 / 3} \paren {\map \cos {\dfrac {\theta + 2 \pi k} 3} + i \, \map \sin {\dfrac {\theta + 2 \pi k} 3} }: k \in \set {0, 1, 2} }$
There are $3$ distinct such complex cube roots.
These can also be expressed as:
:$z^{1 / 3} = \set {r^{1 / 3} e^{i \paren {\theta + 2 \pi k} / 3}: k \in \set {0, 1, 2} }$
or:
:$z^{1 / 3} = \set {r^{1 / 3} e^{i \theta / 3} \omega^k: k \in \set {0, 1, 2} }$
where $\omega = e^{2 i \pi / 3} = -\dfrac 1 2 + \dfrac {i \sqrt 3} 2$ is the first cube root of unity.
== Proof ==
An example of Roots of Complex Number.
== Sources ==
* : Introduction
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of -1+i

== Example of Roots of Complex Number: Corollary ==
The complex cube roots of $-1 + i$ are given by:
:$\paren {-1 + i}^{1/3} = \set {2^{1/6} \paren {\cos \dfrac \pi 4 + i \sin \dfrac \pi 4}, 2^{1/6} \paren {\cos \dfrac {11 \pi} {12} + i \sin \dfrac {11 \pi} {12} }, 2^{1/6} \paren {\cos \dfrac {19 \pi} {12} + i \sin \dfrac {19 \pi} {12} } }$
== Proof ==
:
Let $z^3 = -1 + i$.
We have that:
:$z^3 = \sqrt 2 \paren {\map \cos {\dfrac {3 \pi} 4 + 2 k \pi} + i \, \map \sin {\dfrac {3 \pi} 4 + 2 k \pi} }$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Roots of Complex Numbers: $29 \ \text {(a)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of -11-2i

== Example of Roots of Complex Number: Corollary ==
The complex cube roots of $-11 - 2 i$ are given by:
:$\paren {-11 - 2 i}^{1/3} = \set {1 + 2 i, -\dfrac 1 2 + \sqrt 3 + \paren {-1 - \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {-1 + \dfrac {\sqrt 3} 2} i}$
== Proof ==
Let $z^3 = -11 - 2 i = \paren {p + iq}^3$.
Then:
From this we have:
and:
Thus:
Let $w = \dfrac p q$:
Putting $\dfrac p q = \dfrac 1 2$ leads to::
:$2 p = q$
and hence:
So this gives:
:$z = \begin {cases} 1 + 2i \\ -\dfrac 1 2 + \sqrt 3 + i \paren {-1 - \dfrac {\sqrt 3} 2} \\ -\dfrac 1 2 - \sqrt 3 + i \paren {-1 + \dfrac {\sqrt 3} 2} \end{cases}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $99$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of -11-2i/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: Roots of Complex Numbers: $99$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:Find the cube roots of $-11 - 2 i$.
:Ans. ... $1 + 2 i, \dfrac 1 2 - \sqrt 3 + \paren {1 + \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {\dfrac {\sqrt 3} 2 - 1} i$
== Correction ==
The correct solution is:
:$\set {1 + 2 i, -\dfrac 1 2 + \sqrt 3 + \paren {-1 - \dfrac {\sqrt 3} 2} i, -\dfrac 1 2 - \sqrt 3 + \paren {-1 + \dfrac {\sqrt 3} 2} i}$
as demonstrated in Cube Roots of $-11 - 2 i$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $99$
Category:Mistakes/Theory and Problems of Complex Variables
Roots of Complex Number/Examples/Cube Roots of 2+11i

== Example of Roots of Complex Number ==
The complex cube roots of $2 + 11 i$ are given by:
:$\paren {2 + 11 i}^{1/3} = \set {2 + i, -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3}}$
== Proof ==
Let $z^3 = 2 + 11 i = \paren {p + iq}^3$.
Then:
From this we have:
and:
Thus:
Let $w = \dfrac p q$:
Putting $\dfrac p q = 2$ leads to::
:$p = 2q$
and hence:
So this gives:
:$z = \begin {cases} 2 + i \\ -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 } \\ -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3} \end{cases}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of 2+2i

== Example of Roots of Complex Number: Corollary ==
The complex cube roots of $2 + 2 i$ are given by:
:$\paren {2 + 2 i}^{1/3} = \set {\sqrt 2 \paren {\cos \dfrac \pi {12} + i \sin \dfrac \pi {12} }, -1 + i, -\sqrt 2 \paren {\cos \dfrac {5 \pi} {12} + i \sin \dfrac {5 \pi} {12} }}$
== Proof ==
Let $z = 2 + 2 i$.
Then:
Then:
and so:
:$\arg z = \dfrac \pi 4$
Let $b$ be defined as:
Then we have that the complex cube roots of unity are:
:$1, \exp {\dfrac {2 i \pi} 3}, \exp {\dfrac {-2 i \pi} 3}$
Thus from Roots of Complex Number: Corollary:
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $3$.
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of 2-11i

== Example of Roots of Complex Number ==
The complex cube roots of $2 - 11 i$ are given by:
:$\paren {2 - 11 i}^{1/3} = \set {2 - i, -1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3}}$
== Proof ==
Let $z^3 = 2 - 11 i = \paren {p + iq}^3$.
Then:
From this we have:
and:
Thus:
Let $w = \dfrac p q$:
Putting $\dfrac p q = -2$ leads to::
:$p = -2q$
and hence:
So this gives:
:$z = \begin {cases} 2 - i \\ -1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 } \\ -1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3} \end{cases}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of 2 + 2 root 3 i

== Example of Roots of Complex Number ==
The complex cube roots of $2 + 2 \sqrt 3 i$ are given by:
:$\paren {2 + 2 \sqrt 3 i}^{1/3} = \set {\sqrt [3] 4 \, \map \cis {20 + 120 k} \degrees}$
for $k = 0, 1, 2$.
That is:
== Proof ==
:
Let $z^3 = 2 + 2 \sqrt 3 i$.
We have that:
:$z^3 = 4 \, \map \cis {\dfrac \pi 3 + 2 k \pi} = 4 \, \map \cis {60 \degrees + k \times 360 \degrees}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(c)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Cube Roots of 8

== Example of Roots of Complex Number ==
The complex cube roots of $8$ are given by:
:$\paren 8^{1/3} = \set {2 \, \map \cis {120 k} \degrees}$
for $k = 0, 1, 2$.
That is:
== Proof ==
:
Let $z^3 = 8$.
We have that:
:$z^3 = 8 \, \map \cis {0 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(a)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Square Roots of 2 root 3 - 2i

== Example of Roots of Complex Number ==
The complex square roots of $2 \sqrt 3 - 2 i$ are given by:
:$\paren {2 \sqrt 3 - 2 i}^{1/2} = \set {2 \cis 165 \degrees, 2 \cis 345 \degrees}$
== Proof ==
:
Let $z^2 = 2 \sqrt 3 - 2 i$.
We have that:
:$z^2 = 4 \paren {\dfrac {\sqrt 3} 2 - \dfrac 1 2 i}$
and it is seen (indirectly) from Cube Roots of Unity that:
:$\dfrac {\sqrt 3} 2 - \dfrac 1 2 i = \cis \dfrac {11 \pi} 6$
Hence 
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(a)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Square Roots of 4 root 2 + 4 root 2 i

== Example of Roots of Complex Number ==
The complex square roots of $4 \sqrt 2 + 4 \sqrt 2 i$ are given by:
:$\paren {4 \sqrt 2 + 4 \sqrt 2 i}^{1/2} = \set {\sqrt 8 \cis 22.5 \degrees, \sqrt 8 \cis 202.5 \degrees}$
== Proof ==
:
Let $z^2 = 4 \sqrt 2 + 4 \sqrt 2 i$.
We have that:
:$z^2 = 8 \paren {\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i}$
and it is seen that:
:$\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i = \cis \dfrac \pi 4$
Hence 
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $96 \ \text{(b)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/Square of Cube Roots of i

== Example of Roots of Complex Number: Corollary ==
The squares of the complex cube roots of $i$ are given by:
:$\paren i^{2/3} = \set {\map \cis {60 + 120 k} \degrees}$
for $k = 0, 1, 2$.
That is:
== Proof ==
:
Let $z^3 = i$.
We have that:
:$z^3 = \map \cis {90 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $95 \ \text{(f)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/z^4 + 81 = 0

== Theorem ==
The roots of the polynomial:
:$z^4 + 81$
are:
:$\set {3 \cis 45 \degrees, 3 \cis 135 \degrees, 3 \cis 225 \degrees, 3 \cis 315 \degrees}$
== Proof ==
From Factorisation of $z^n + 1$:
:$z^4 + 1 = \ds \prod_{k \mathop = 0}^3 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 4}$
Thus:
:$z = \set {\exp \dfrac {\paren {2 k + 1} i \pi} 4}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(a)}$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/z^4 - 81 = 0

== Theorem ==
The roots of the polynomial:
:$z^4 - 81$
are:
:$\set {3, 3 i, -3, -3 i}$
== Proof ==
From Factorisation of $z^n - a$:
:$z^4 - a = \ds \prod_{k \mathop = 0}^3 \paren {z - \alpha^k b}$
where:
:$\alpha$ is a primitive complex $4$th root of unity
:$b$ is any complex number such that $b^4 = a$.
Here we can take $b = 3$, as $81 = 3^4$.
Thus:
:$z = \set {3 \exp \dfrac {k i \pi} 2}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Ponderable $2.1.2$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/z^5 + 1 = 0

== Theorem ==
The roots of the polynomial:
:$z^5 + 1 = 0$
are:
:$\set {\cos \dfrac \pi 5 \pm i \sin \dfrac \pi 5, \cos \dfrac {3 \pi} 5 \pm i \sin \dfrac {3 \pi} 5, -1}$
== Proof ==
From Factorisation of $z^n + 1$:
:$z^5 + 1 = \ds \prod_{k \mathop = 0}^4 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 5}$
Thus:
:$z = \set {\exp \dfrac {\paren {2 k + 1} i \pi} 5}$
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $9$
Category:Examples of Complex Roots
Roots of Complex Number/Examples/z^6 + 1 = root 3 i

== Theorem ==
The roots of the equation:
:$z^6 + 1 = \sqrt 3 i$
are:
:$\set {\sqrt [6] 2 \cis 20 \degrees, \sqrt [6] 2 \cis 80 \degrees, \sqrt [6] 2 \cis 140 \degrees, \sqrt [6] 2 \cis 200 \degrees, \sqrt [6] 2 \cis 260 \degrees, \sqrt [6] 2 \cis 320 \degrees}$
== Proof ==
We have:
:$z^6 = -1 + \sqrt 3 i$
and so this is equivalent to finding the complex $6$th roots of $-1 + \sqrt 3 i$:
We have that:
:$z^6 = 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}$
Let $z = r \cis \theta$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Roots of Complex Numbers: $97 \ \text{(b)}$
Category:Examples of Complex Roots