Examples On Roots Of Complex Numbers
Example - 26
Find the square roots and the cube roots of i.
Solution: First write i in its Euler’s form.
\[\begin{align}&i = {e^{\frac{{i\pi }}{2}}}\end{align}\]
Square roots:
\[\begin{align}&{z^2} = {e^{i\pi /2}} = {e^{i\left( {2p\pi + \frac{\pi }{2}} \right)}}\\ &\;\;\;\;\;\Rightarrow \,\,\,z = {e^{i\left( {p\pi + \frac{\pi }{4}} \right)}}\end{align}\]
The two roots are given by two consecutive values of p, say p = 0, 1
\[ \begin{align}&\Rightarrow \,\,\,{z_1} = {e^{i\,\,\pi /4}},\,\,{z_2} = {e^{i\,\,5\pi /4}}\end{align}\]
Cube roots:
\[\begin{align}&\;\;\;\;\;\;{z^3} = {e^{i\left( {2p\pi + \frac{\pi }{2}} \right)}}\\ &\Rightarrow \,\,\,z = {e^{i\left( {\frac{{2p\pi }}{3} + \frac{\pi }{6}} \right)}} \end{align}\]
There are three roots; take \[\begin{align}&\Rightarrow \,\,\,{z_1} = {e^{i\,\pi /6}},\,\,{z_2} = {e^{i5\,\pi /6}},\,\,{z_3} = {e^{i3\,\pi /2}}\end{align}\]
Let us plot the square roots and cube roots of i on the plane:
Example - 27
If \(1,\,\,{\alpha _1},\,\,{\alpha _2}...,\,\,{\alpha _{n - 1}}\) are the \({n^{th}}\) roots of unity, find the value of \((1 + {\alpha _1})(1 + {\alpha _2})...(1 + {\alpha _{n - 1}})\) .
Solution: Since \({x^n} - 1 = 0\) has n roots (that have been specified in the question), we can write (by factor theorem):
\[\begin{align}&\qquad\;\;{x^n} - 1 = (x - 1)(x - {\alpha _1})...(x - {\alpha _{n - 1}})\\&\Rightarrow \,\,\,\frac{{{x^n} - 1}}{{x - 1}} = (x - {\alpha _1})(x - {\alpha _2})...(x - {\alpha _{n - 1}})\;\;\;\;\;\;\;....(1)\end{align}\]
The form of the expression whose value we need to obtain hints that we should substitute \(x = - 1\) in (1):
\[\begin{align}&\frac{{{{( - 1)}^n} - 1}}{{( - 1) - 1}} = ( - 1 - {\alpha _1})( - 1 - {\alpha _2})...( - 1 - {\alpha _{n - 1}})\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\quad\;\;= {( - 1)^{n - 1}}(1 + {\alpha _1})(1 + {\alpha _2})...(1 + {\alpha _{n - 1}})\end{align}\]
If n is even the left hand side is 0 so that the value of the expression becomes 0. If n is odd, the expression becomes
\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{ - 2}}{{ - 2}} = {( - 1)^{n - 1}}(1 + {\alpha _1})(1 + {\alpha _2})...(1 + {\alpha _{n - 1}})\\ &\;\;\;\Rightarrow \,\,\,(1 + {\alpha _1})(1 + {\alpha _2})...(1 + {\alpha _{n - 1}}) = 1\end{align}\]
Thus, the expression takes the value 0 or 1 depending on whether n is even or odd respectively.
Example - 28
Let \(\omega ,\,\,{\omega ^{\,2}}\) be the complex cube roots of unity. Let \({z_{1,}}\,\,{z_2},\,\,{z_3}\) be complex numbers such that
\[\begin{array}{l}\,\,\,\,\,\,\,\,\,{z_1} + {z_2} + {z_3} = A\\{z_1} + \omega {z_2} + {\omega ^2}{z_3} = B\\{z_1} + {\omega ^2}{z_2} + \omega {z_3} = C\end{array}\]
Evaluate \({z_1},\,\,{z_2}{\rm{ and }}{z_3}\) in terms of A, B, and C.
Solution: We know that \(1 + \omega + {\omega ^2} = 0\)
We have to use this to somehow express each of \({z_1},\,\,{z_2}{\rm{ and }}{z_3}\) independently in terms of A, B and C. Label the three equations as (I), (II) and (III). (I) + (II) + (III) gives
\[\begin{align}& \,\,\,\,\,\,\,\,\,\,\,\,3{z_1} = A + B + C\\ &\;\;\Rightarrow \,\,\,{z_1} = \frac{{A + B + C}}{3} \end{align}\]
\(\left( I \right){\rm{ }} + w{^2}{\rm{ }}\left( {II} \right){\rm{ }} + w\left( {III} \right){\rm{ }}gives\)
\[\begin{align}&\;\;\,\,\,\,\,\,3{z_2} = A + {\omega ^2}B + \omega C\\ &\Rightarrow \,\,\,{z_2} = \frac{{A + B{\omega ^2} + C\omega }}{3}\end{align}\]
Finally, \(\left( I \right){\rm{ }} + w{^2}{\rm{ }}\left( {II} \right){\rm{ }} + w\left( {III} \right){\rm{ }}gives\)
\[\begin{align}&\;\,\,\,\,\,\,3{z_3} = A + B\omega + C{\omega ^2}\\ &\Rightarrow \,\,\,{z_3} = \frac{{A + B\omega + C{\omega ^2}}}{3}\end{align}\]
These were the values we were looking for.
Example - 29
Let a complex number \(\alpha ,\,\,\alpha \ne 1\) be a root of \({z^{p + q}} - {z^p} - {z^q} + 1 = 0,\) where p and q are distinct primes. Show that either \(1 + \alpha + {\alpha ^2} + ... + {\alpha ^{p - 1}} = 0\) or \(1 + \alpha + {\alpha ^2} + ... + {\alpha ^{q - 1}} = 0\) , but not both together
Solution: The given equation can be written as
\[({z^p} - 1)({z^q} - 1) = 0\]
Therefore, \(\alpha \) is a root of either \({z^p} - 1 = 0\,\,{\rm{or }}{z^q} - 1 = 0\) . In other words, \(\alpha \) is either a \({p^{th}}\) or a \({q^{th}}\) root of unity.
Now,
\({z^p} - 1 = (z - 1)({z^{p - 1}} + {z^{p - 2}} + .... + z + 1)\)
Substituting \(z = \alpha \) gives
\[0 = (\alpha - 1)({\alpha ^{p - 1}} + {\alpha ^{p - 2}} + ..... + \alpha + 1)\]
Since \(\alpha \ne 1,\) we get
\[1 + \alpha + {\alpha ^2} + .... + {\alpha ^{p - 1}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]
Similarly, if \({z^q} - 1 = 0,\) we get
\[1 + \alpha + {\alpha ^2} + ..... + {\alpha ^{q - 1}} = 0\,\,\, \ldots (2)\]
We now need to show that (1) and (2) cannot hold simultaneously. In other words, \(\alpha\) cannot be a \(p{^{th}}{\rm{ }}\;and\;{\rm{ }}a\;{q^{th}}\) root of unity at the same time (given the condition that p and q are distinct primes). This is easy to prove:
If \(\alpha\) is a p th root of unity,
\[\alpha = {e^{^{i\frac{{2m\pi }}{p}}\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (3)\]
If \(\alpha\) is a q th root of unity,
\[\alpha = {e^{i\frac{{2n\pi }}{q}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 4 \right)\]
Observe the right hand sides of (3) and (4) carefully. The terms in the exponents \(\frac{m}{p}\) , (where m < p) and \(\frac{n}{q}\) (where n < q) can never be equal. Why? Lets assume they are equal
\[\begin{align}&\,\,\,\,\,\,\,\,\,\frac{m}{p} = \frac{n}{q}\\ &\Rightarrow pn = mq\end{align}\]
Since p is a prime, different from q, it cannot have m and q and factors. Since n < q, n cannot have q as a factor. Thus pn cannot have m or q as factors which is a contradiction.
\(\Rightarrow \) (3) and (4) cannot be equal
\(\Rightarrow \) (1) and (2) cannot be simultaneously satisfied.
Here’s an example, for p = 3 and q = 5:
p=3,q=5
Example - 30
Find all the roots of the equation \({z^{12}} - 56{z^6} - 512 = 0\) whose imaginary parts are non-negative
Solution: We let \({z^6} = x\) so that the given equation reduces to a quadratic.
\[\begin{array}{l}\,\,\,\,\,\,\,\,\,{x^2} - 56x - 512 = 0\\\,\,\,\,\,\,\,\,\,(x - 64)(x + 8) = 0\\ \Rightarrow x = 64, - 8\\ \Rightarrow {z^6} = 64, - 8\end{array}\]
(a) \(\fbox{\({z^6}\) = 64}\)
\[\begin{align}&{z^6} = 64 = {2^6}{e^{i(2p\pi + 0)}}\\ &\quad\;\Rightarrow z = 2{e^{i\left( {\frac{{2p\pi + 0}}{6}} \right)}}\end{align}\]
Verify that for p = 0, 1, 2, 3, we get non-negative imaginary parts for z:
\[z = 2,\,\,\,\,2{e^{i\pi {\rm{/3}}}},\,\,\,\,2{e^{i2\pi {\rm{/3}}}},\,\,\,\,\,2{e^{i\pi }}\]
(b) \(\fbox{\({z^6}\) = - 8}\)
\[\begin{array}{l}\,\,\,\,\,\,{z^6} = - 8 = {2^3}{e^{i\pi }} = {2^3}{e^{i(2p\pi + \pi )}}\\ \Rightarrow z = {2^{1{\rm{/2}}}}{e^{i\left( {\frac{{2p\pi + \pi }}{6}} \right)}}\end{array}\]
Verify that for p = 0, 1, 2, we get non-negative imaginary parts for z:
\[\begin{align}&z = \sqrt 2 {e^{i\pi {\rm{/6}}}},\sqrt 2 {e^{i\pi {\rm{/2}}}},\sqrt 2 {e^{i5\pi {\rm{/6}}}}\end{align}\]
Thus we get seven values of z that satisfy the given condition.
TRY YOURSELF - VIII
Q. 1 By considering the eleventh roots of unity, find the value of
\[S = \sum\limits_{k = 1}^{10} {} \left( {\sin \left( {\frac{{2\pi k}}{{11}}} \right) - i\cos \left( {\frac{{2\pi k}}{{11}}} \right)} \right)\]
Q. 2 Find the value of \((\alpha - 1)(\alpha - {\alpha ^2})(\alpha - {\alpha ^3})...(\alpha - {\alpha ^{n - 1}})\) where \(\alpha = {e^{i2\pi /n}}\) .
Q. 3 If ABC is an equilateral triangle having vertices \({z_1}{\rm{ }},{z_2}{\rm{ }}\;and\;{z_3}{\rm{ }}\) (in the anti-clockwise direction), find the value of \({z_1} + \omega {z_2} + {\omega ^2}{z_3}\) .
Q. 4 Find all non-zero solutions of \({z^5} = \bar z\) .
Hint: Use the polar form for z.
Q. 5 Prove that the following factorizations hold true:
(a) \({x^2} + x + 1 = (x - \omega )(x - {\omega ^2})\)
(b) \({x^2} - x + 1 = (x + \omega )(x + {\omega ^2})\)
(c) \({x^2} + xy + {y^2} = (x - y\omega )(x - y{\omega ^2})\)
(d) \({x^3} + {y^3} = (x + y)(x + y\omega )(x + y{\omega ^2})\)
(e) \({x^3} - {y^3} = (x - y)(x - y\omega )(x - y{\omega ^2})\)
(f) \({x^2} + {y^2} + {z^2} - xy - yz - zx = (x + y\omega + z{\omega ^2})(x + {\omega ^2}y + \omega z)\)