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.

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

(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.)

Key Idea

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

Step 1

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.

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 \( (2^{n/2})^2 = 2^n \)