Question

Assertion :If $z_0=\frac{1}{2}(1+i)$, then $P_n\left(z\right)=(1+z_0)(1+{z_0}^2)(1+{z_0}^{2^2})...(1+{z_0}^{2^n})=(1+i)(1-\frac{1}{2^{2^n}})$, where $n>1$ is a positive integer. Reason: $P_n\left(z\right)=\frac{1-{z_0}^{2^{n+1}}}{1-z_0}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We have, $P_n\left(z\right)=(1+z_0)(1+{z_0}^2)(1+{z_0}^{2^2})...(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P_n\left(z\right)=(1-z_0)(1+z_0)(1+{z_0}^2)(1+{z_0}^{2^2})...(1+{z_0}^{2^n})=(1-{z_0}^2)(1+{z_0}^2)(1+{z_0}^{2^2})...(1+{z_0}^{2^n})=(1-{z_0}^{2^2})(1+{z_0}^{2^2})...(1+{z_0}^{2^n})=...=(1-{z_0}^{2^{n+1}})$

Now ${z_0}^2=\frac{i}{2}\Rightarrow {z_0}^{2^{n+1}}={\left({z_0}^2\right)}^{2^n}={\left(\frac{i}{2}\right)}^{2^n}=\frac{i^{2^n}}{2^{2^n}}$

now, since $2_n$ is divisible by $4,$ if $n>1\Rightarrow i^{2^n}=1$

Thus, $1-{z_0}^{2^{n+1}}=1-\frac{1}{2^{2^n}}\Rightarrow P_n\left(z\right)=\frac{1}{1-z_0}(1-\frac{1}{2^{2^n}})=(1+i)(1-\frac{1}{2^{2^n}})$