Question

If ${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$ , where $r = 1,2,3....n$ then $\mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right)$ is equal to
A) $\cos \alpha + i\sin \alpha$
B) $\cos \alpha + i\sin \alpha$
C) ${e^{\dfrac{{i\alpha }}{2}}}$
D) $\sqrt {{e^{i\alpha }}}$

Answer 1

Here we need to find the value of the given expression. We will first use the Euler’s formula for the given complex number and write it in the exponential form. Then we will put the values of the variable in the complex number and substitute it in the given expression. We will simplify it to get the required value of the given expression.

Complete step by step solution:
We know the Euler’s formula for the complex number is ${e^{i\theta }} = \cos \theta + i\sin \theta$.
Now, we will apply this formula to a given complex number. Therefore, we get
${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}} = {e^{i\dfrac{{r\alpha }}{{{n^2}}}}}$
$\Rightarrow {z_r} = {e^{i\dfrac{{r\alpha }}{{{n^2}}}}}$ …………. $\left( 1 \right)$
Now, we will find the value of ${z_1}$ by substituting the value of $r$ as 1. Therefore, we get
${z_1} = {e^{\dfrac{{i\alpha }}{{{n^2}}}}}$
Similarly, we will find the value of ${z_2}$ substituting the value of $r$ as 2. So,
${z_2} = {e^{\dfrac{{i2\alpha }}{{{n^2}}}}}$
Similarly, we get a generalized equation as:
${z_n} = {e^{\dfrac{{in\alpha }}{{{n^2}}}}}$
But here we have to calculate the value of $\mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right)$.
Now, we will substitute all these values in the given expression.
$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {{e^{\dfrac{{i\alpha }}{{{n^2}}}}} \cdot {e^{\dfrac{{i2\alpha }}{{{n^2}}}}} \cdot {e^{\dfrac{{i3\alpha }}{{{n^2}}}}}.......{e^{\dfrac{{in\alpha }}{{{n^2}}}}}} \right)$
We know that when we multiply the exponents with the same base, their powers get added.
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {{e^{\dfrac{{i\alpha }}{{{n^2}}} + }}{{^{\dfrac{{i2\alpha }}{{{n^2}}} + }}^{\dfrac{{i3\alpha }}{{{n^2}}} + .......}}^{ + \dfrac{{in\alpha }}{{{n^2}}}}} \right)$
On further simplifying the exponent, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = \mathop {\lim }\limits_{n \to \infty } {e^{\dfrac{{i\alpha }}{{{n^2}}}\left( {1 + 2 + 3 + ..... + n} \right)}}$
We know the formula sum of series $1 + 2 + 3 + .... + n$ is equal to $\dfrac{{n\left( {n + 1} \right)}}{2}$ .
Now, using the formula in the exponent of the above question, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = \mathop {\lim }\limits_{n \to \infty } {e^{\dfrac{{i\alpha }}{{{n^2}}} \times \dfrac{{\left( {n + 1} \right)n}}{2}}}$
On multiplying the exponents, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = \mathop {\lim }\limits_{n \to \infty } {e^{\dfrac{{i\alpha }}{2}\left( {1 + \dfrac{1}{n}} \right)}}$
On applying the limits, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = {e^{\dfrac{{i\alpha }}{2}\left( {1 + 0} \right)}} = {e^{\dfrac{{i\alpha }}{2}}}$
On further simplifying the exponents, we get
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{z_1} \cdot {z_2}......{z_n}} \right) = {e^{i\alpha \times \dfrac{1}{2}}} = \sqrt {{e^{i\alpha }}}$

Hence, the correct option is option D.

Note:
To solve such types of problems, we need to know the basic properties of limits and also the exponentials. When two or more exponents with the same base are multiplied then their powers get added. Similarly, when we divide one exponential by another exponential of the same base then their powers are subtracted. We might make a mistake by applying the limit in the beginning as this gives us an answer as infinity. Therefore, we must apply limits at the end of the calculation.