Question

If $z+{{z}^{-1}}=1$ then find the value of ${{z}^{100}}+{{z}^{-100}}$.

Answer 1

Hint: In this question, we are given the value of $z+{{z}^{-1}}=1$ and we need to find the value of ${{z}^{100}}+{{z}^{-100}}$. For this, we will first find the values of z from $z+{{z}^{-1}}=1$ which will be complex numbers. Then we will compare them to the cube root of unity and find values of z in terms of cube root of unity. Then we will put values in ${{z}^{100}}+{{z}^{-100}}$ and further evaluate using properties of the cube root of unity. Cube roots of units are given as $w=\dfrac{-1+i\sqrt{3}}{2}\text{ and }{{w}^{2}}=\dfrac{-1-i\sqrt{3}}{2}$. Properties of cube roots of unity that we will use are:
(i) ${{w}^{3}}=1$.
(ii) Sum of cube roots of unity $1+w+{{w}^{2}}=0$.

Complete step by step answer:
Here we are given that $z+{{z}^{-1}}=1$.
${{z}^{-1}}$ can be written as $\dfrac{1}{z}$ so we get, $z+\dfrac{1}{z}=1$.
Taking LCM as z, we get, $\dfrac{{{z}^{2}}+1}{z}=1$.
Cross multiplying we get, ${{z}^{2}}+1=z\Rightarrow {{z}^{2}}-z+1=0$.
Now let us find the value of z using the quadrant formula. Quadratic formula for an equation $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
For ${{z}^{2}}-z+1=0$ we have,
\[\begin{align}
  & z=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow \dfrac{1\pm \sqrt{1-4}}{2} \\
 & \Rightarrow \dfrac{1\pm \sqrt{3}i}{2} \\
 & \Rightarrow \dfrac{1+\sqrt{3}}{2}\text{ and }\dfrac{1-\sqrt{3}}{2} \\
\end{align}\]
We know that cube roots of unity are given as $w=\dfrac{-1+i\sqrt{3}}{2}\text{ and }{{w}^{2}}=\dfrac{-1-i\sqrt{3}}{2}$.
So, $z=\dfrac{1+\sqrt{3}i}{2}$ can be written as $z=-\left( \dfrac{-1-\sqrt{3}i}{2} \right)=-{{w}^{2}}$ and $z=\dfrac{1-\sqrt{3}i}{2}$ can be written as \[z=-\left( \dfrac{-1+\sqrt{3}i}{2} \right)=-w\].
Hence the values of z are $-w\text{ and }-{{w}^{2}}$.
Now let us find the value of ${{z}^{100}}+{{z}^{-100}}$ using cube roots of units.
Putting z = -w, we get ${{\left( -w \right)}^{100}}+{{\left( -w \right)}^{-100}}$.
As we know, ${{a}^{-1}}=\dfrac{1}{a}$ so we get, ${{\left( -w \right)}^{100}}+\dfrac{1}{{{\left( -w \right)}^{100}}}$.
We know that, ${{\left( -a \right)}^{2n}}={{\left( a \right)}^{2n}}$ and 100 is multiple of 2 therefore, we get, ${{\left( w \right)}^{100}}+\dfrac{1}{{{\left( w \right)}^{100}}}$.
Now, we know that ${{w}^{3}}=1$.
So to calculate ${{w}^{100}}$ we can first calculate ${{w}^{99}}$ which is a multiple of 3.
Taking power of 33 on both sides of ${{w}^{3}}=1$ we get: ${{\left( {{w}^{3}} \right)}^{33}}={{\left( 1 \right)}^{33}}\Rightarrow {{w}^{99}}=1$.
We can write ${{w}^{100}}={{w}^{99}}\cdot w$ but ${{w}^{99}}=1$ hence, we get, ${{w}^{100}}=1\cdot w=w$.
So our expression become ${{w}^{100}}+\dfrac{1}{{{w}^{100}}}=w+\dfrac{1}{w}$.
Now let us take LCM of w, we get: $\dfrac{{{w}^{2}}+1}{w}$.
From the sum of cube roots of unity, we know that $1+w+{{w}^{2}}=0$. Therefore we can say that $1+{{w}^{2}}=-w$.
Putting it in the above expression we get: $\dfrac{-w}{w}$.
Cancelling w, we get: -1
Hence by putting z = -w, we get the value of ${{z}^{100}}+{{z}^{-100}}$ as -1.
Now, let us put value of $z=-{{w}^{2}}$ we get:
${{z}^{100}}+\dfrac{1}{{{z}^{100}}}={{\left( -{{w}^{2}} \right)}^{100}}+\dfrac{1}{{{\left( -{{w}^{2}} \right)}^{100}}}={{w}^{200}}+\dfrac{1}{{{w}^{200}}}$
Now, we know that nearest multiple of 3 around 200 is 198 and we know that ${{w}^{3}}=1$ so taking power of 66 on both sides of ${{w}^{3}}=1$ we get: \[{{\left( {{w}^{3}} \right)}^{66}}={{1}^{66}}\Rightarrow {{w}^{198}}=1\].
Now we can write ${{w}^{200}}$ as \[{{w}^{198}}\cdot {{w}^{2}}\]. Hence ${{w}^{200}}=1\cdot {{w}^{2}}={{w}^{2}}$ so our expression becomes \[{{w}^{2}}+\dfrac{1}{{{w}^{2}}}\].
Now we know that, ${{w}^{3}}=1$ we can write it as $w\left( {{w}^{2}} \right)=1$.
Taking ${{w}^{2}}$ to the other side, we get $w=\dfrac{1}{{{w}^{2}}}$.
Hence putting the value of $\dfrac{1}{{{w}^{2}}}$ in the above expression we get: ${{w}^{2}}+w$.
We know the sum of cube roots of unity $1+w+{{w}^{2}}=0$ so we can write $w+{{w}^{2}}=-1$.
Therefore, our expression becomes equal = -1
Hence by putting $z=-{{w}^{2}}$ we get value of ${{z}^{100}}+{{z}^{-100}}$ as -1.

Hence -1 is our required value.

Note: Students should take care of the signs while solving the equation and finding values. Always try to find values of ${{z}^{100}}+{{z}^{-100}}$ using both the obtained values of z. Students should keep in mind the properties and the values of cube roots of unity to solve these questions.