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Explore the Back-StoryThis is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

If \(a_0, a_1, \cdots, a_n \) are real numbers such that $$ (1+z)^n = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n $$ for all complex numbers z, then the value of $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$ equals

(A) \( 2^n \) ; (B) \( a_0^2 + a_1^2 + \cdots + a_n^2 \) ; (C) \( 2^{n^2} \) ; (D) \( 2n^2 \) ;

(*How to use this discussion:* **Do not read the entire solution at one go. First, **read more on the **Key Idea, **then give the problem a try. **Next, **look into **Step 1 **and give it another try and so on.)

This is the generic use case of Complex Number \( \iota =\sqrt {-1} \) and binomial theorem.

Note that \( i^2 = -1 \). Also, geometrically speaking, i = (0,1). Hence adding (1,0) to i (=(0,1)) gives us the point (1, 1). Polar coordinate of this point is \( ( \sqrt 2, \frac{pi}{4} ) \). Here is a picture:

Try the problem with this hint before looking into step 2. Remember, no one learnt mathematics by looking at solutions.

**At Cheenta we are busy with Complex Number and Geometry module. Additionally I.S.I. Entrance Mock Test 1 is also active now.**

Replace \( z \) by \( i \). We have \( (1+z)^n = (\sqrt 2 , \frac {\pi}{4} )^n = (2^{n/2}, \frac{n \cdot \pi }{4} ) \) on the left hand side.

Now, replace \( z \) by \( i \) on the right hand side.

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Replacing z by \( i \) on the right hand side we have $$(2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 + a_1 i + a_2 i^2 + a_3 I^3 \cdots + a_n I^n $$. This implies $$ (2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 - a_2 + a_4 - \cdots + i (a_1 - a_3 + a_5 - \cdots ) $$

**Think** now, what the following expression represents: $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$

It represents the **square of the** length of point \( (2^{n/2}, \frac{n \cdot \pi }{4} ) \). That is simply \( (2^{n/2})^2 = 2^n \)

Hence the answer is option A.

Look into the following notes on Complex Number and Geometry module.

This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

If \(a_0, a_1, \cdots, a_n \) are real numbers such that $$ (1+z)^n = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n $$ for all complex numbers z, then the value of $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$ equals

(A) \( 2^n \) ; (B) \( a_0^2 + a_1^2 + \cdots + a_n^2 \) ; (C) \( 2^{n^2} \) ; (D) \( 2n^2 \) ;

(*How to use this discussion:* **Do not read the entire solution at one go. First, **read more on the **Key Idea, **then give the problem a try. **Next, **look into **Step 1 **and give it another try and so on.)

This is the generic use case of Complex Number \( \iota =\sqrt {-1} \) and binomial theorem.

Note that \( i^2 = -1 \). Also, geometrically speaking, i = (0,1). Hence adding (1,0) to i (=(0,1)) gives us the point (1, 1). Polar coordinate of this point is \( ( \sqrt 2, \frac{pi}{4} ) \). Here is a picture:

Try the problem with this hint before looking into step 2. Remember, no one learnt mathematics by looking at solutions.

**At Cheenta we are busy with Complex Number and Geometry module. Additionally I.S.I. Entrance Mock Test 1 is also active now.**

Replace \( z \) by \( i \). We have \( (1+z)^n = (\sqrt 2 , \frac {\pi}{4} )^n = (2^{n/2}, \frac{n \cdot \pi }{4} ) \) on the left hand side.

Now, replace \( z \) by \( i \) on the right hand side.

[/tab]

[tab]

Replacing z by \( i \) on the right hand side we have $$(2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 + a_1 i + a_2 i^2 + a_3 I^3 \cdots + a_n I^n $$. This implies $$ (2^{n/2}, \frac{n \cdot \pi }{4} ) = a_0 - a_2 + a_4 - \cdots + i (a_1 - a_3 + a_5 - \cdots ) $$

**Think** now, what the following expression represents: $$ (a_0 - a_2 + a_4 - a_6 + \cdots )^2 + (a_1 - a_3 + a_5 - a_7 + \cdots )^2 $$

It represents the **square of the** length of point \( (2^{n/2}, \frac{n \cdot \pi }{4} ) \). That is simply \( (2^{n/2})^2 = 2^n \)

Hence the answer is option A.

Look into the following notes on Complex Number and Geometry module.

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