Edition 11: IIT JEE Complex Numbers

Problem 1 : Euler's equation and its implications
Don't get nervous because you haven't heard of this term. It is actually quite simple.
You have learnt in infinite series, that
Sin(x) = x x3/3! + x5/5! x7/7! + …….
And
Cos(x) = 1 x2/2! + x4/4! …..
Now let's evaluate the expression
Cos(x) + i. Sin(x)
= 1 + ix + i2 . x2/3 + i3. x3/3! + i4.x4/4! + i5.x5/5! + i6.x6/5! + i7.x7/7! +
We can write the nth term of the above series as
(ix)n / n! … (very yourself) and you realize that it is the exponential series of (i.x)
Therefore
eix = cos(x) + i . sin(x)
The theory of this equation is not really part of IIT JEE syllabus.
Now I would show some tricky applications of this equation
1. Substitute x = π and you get
ei π = -1
Ÿ ei π + 1 = 0
This expression magically connects the 5 most fundamental constants on nature in one simple equation.
2. Continuing with the same
equation, we observe that ei π = i2
or, ii = ei π . I/2 or, ii = e - π/2, which
is a real number, amazing !
3. e3i = cos(3x) + i . Sin(3x) or we can
write (cos x + i sin x ) 3 = cos(3x) + i.
sin(3x), after expanding the left
hand side and then segregating the
imaginary and real parts we can
easily prove the expressions like
cos 3x = cos3x - 3sin2x cos x
Problem 2 : A different view of Simple Harmonic Motion
In the chapters on Simple Harmonic Motion you would learn to solve the time period and other parameters of an oscillating system using the standard equation F(x) = - k.x
The way they teach you to construct such an equation is to identify the net force on the system after imparting a small displacement of x from the equilibrium position and the then equating the same with mass X acceleration or rotational equivalent.
However, in many cases, it is very difficult to identify the net force in terms of displacement x. In such situation we only know the energy content of the system.
Also, we know that for any conservative system, the net energy = Potential (U)+ kinetic = constant.
Ÿ U + ½ m .v.v = constant
Ÿ U = -1/2 . m . (v) 2
Ÿ dU/dx = -1/2 . m . 2. (v). (dv/dx)
Ÿ dU/dx = -m. v. dv/dx = - m. (dx/dt).
[d/dx ( dx/dt) ]
Ÿ dU/dx = - m . d/dt (d/dt (x))
Ÿ dU/dx = - F , where F is the force.
Therefore the equation of SHM can be written as
dU/dx = kx
Ÿ d2U/dx2 = k
In other words, the 'spring constant' can be expressed in terms of potential energy of the system.
Example : To create a soap bubble you need to blow air into to it. The work you do in blowing gets converted to potential energy in the form of surface tension. Although this is not part of IIT JEE syllabus, please note that the surface energy of the bubble = the surface tension of soap bubble (which is a constant given a particular chemical composition of soap water) X the surface area of the bubble. In other words U = S . 4π r2
Where S is the surface tension and r is the radius of the soap bubble. Note that the energy content is positive and by nature, any system tends to lose potential energy. The soap bubble will also tend to reduce in size and will stabilize with some radius R where the air pressure from inside will balance out the force due to shrinking of the bubble.
If you disturb the bubble by denting it gently by blowing air from outside, the geometry of the soap bubble changes and some potential energy gets introduced in to the system. Assuming some amount of simplicity, we can say that the average radius of the soap bubble changes and this causes a simple harmonic motion. To find out the time period of this motion, we observe that U = S . 4π r2
Ÿ d2U/dr2 = 8.S.π = constant and it satisfies the pre-condition of SHM.
Therefore the spring constant is 8.S.π and therefore the time period will be
T = 2 π