Question

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

Answer 1

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,

& 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{aligned}$$ We know that cube roots of unity are given as $w=\dfrac{-1+i\sqrt{3}}{2}\text{ and }{{w}^{2}}=\dfrac{-1-i\sqrt{3}}{2}$. So, $z=\dfrac{1+\sqrt{3}i}{2}$ can be written as $z=-\left( \dfrac{-1-\sqrt{3}i}{2} \right)=-{{w}^{2}}$ and $z=\dfrac{1-\sqrt{3}i}{2}$ can be written as $$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.