Expert: Paul Klarreich - 3/2/2009

Question
Hi, I have this question in which I was able to do up to part (b) i] but am now stuck with the rest. The question is:

(a) Prove, using mathematical induction, that for a positive integer n, (cosX+isinX)^n = cos(nX) + isin (nX) where i^2 = -1. [I have done this part already]

(b)The complex number is defined by z= cosX +isinX.
i] Show that 1/z= cos(-X) + isin(-X)[I have done this already too]
ii] Deduce that z^n + z^-n = 2cos(nX) [This is the part I get stuck at]

(c) i]Find the binomial expansion of (z+z^-1)^5.
ii] Hence, show that (cosX)^5 = 1/16(acos5X + bcos3X + ccosX) where a,b,c are positve integers to be found.

These last parts confuse me greatly. Any help you can provide would be very helpful!!

Answer
Questioner:   Kaela
Category:  Advanced Math
Private:  No
 
Subject:  Complex Numbers confusion
Question:  Hi, I have this question in which I was able to do up to part (b) i] but

am now stuck with the rest. The question is:

(a) Prove, using mathematical induction, that for a positive integer n,

(cosX+isinX)^n = cos(nX) + isin (nX) where i^2 = -1. [I have done this part already]

(b)The complex number is defined by z= cosX +isinX.
i] Show that 1/z= cos(-X) + isin(-X)[I have done this already too]

ii] Deduce that z^n + z^-n = 2cos(nX) [This is the part I get stuck at]
..................................................


OK, then, if you already know that:

1/z = cos(-x) + i sin(-x),

Then, deduce (Oh, those authors love that word, don't they.)

z^-n = (1/z)^n = (cos(-X) + i sin(-X))^n = cos(-nX) + i sin(-nX)

Now use these trig facts:

cos(-A)  =  cos A   [we say cosine is an even function]
sin(-A)  = - sin A   [we say sine is an odd function]

and conclude (another lovely word) that:

z^-n = (1/z)^n = (cos(-X) + i sin(-X))^n = cos(-nX) + i sin(-nX) =
z^-n    = cos(nx) - i sin(nx)

So

z^n   = cos(nx) + i sin(nx), and
z^-n  = cos(nx) - i sin(nx)

I wonder what comes next.  Right:

z^n + z^-n = 2 cos(nx)

(c) i]Find the binomial expansion of (z+z^-1)^5.

Well, the b.e. of

(a + b)^5 =  a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

So put a = z, b = 1/z, and cancel powers when you can, like z^3 * (1/z)^2 = z, right?

z^5 + 5z^3 + 10z + 10/z + 5/z^3 + 1/z^5

...................

ii] Hence, show that (cosX)^5 = 1/16(acos5X + bcos3X + ccosX) where a,b,c are

positve integers to be found.


Take that b.e. and rearrange into three pieces:

[z^5 + 1/z^5] + [5z^3 + 5/z^3] + [10z + 10/z]  << THE THREE PIECES



Now let's make use of that theorem you proved earlier:

z^n + 1/z^n = 2 cos(nx), and use it for:

THE LEFT SIDE:  (z+z^-1)^5 = (2 cos x)^5 = 32 (cos x)^5
THE FIRST PIECE: z^5 + 1/z^5 = 2 cos(5x)
THE SECOND PIECE: 5(z^3 + 1/z^3) = 10 cos(3x)
THE THIRD  PIECE: 10(z + 1/z) = 20 cos(x)

32 (cos x)^5 = 2 cos(5x) + 10 cos(3x) + 20 cos(x)

divide by 2:

16 (cos x)^5 =  cos(5x) + 5 cos(3x) + 10 cos(x)

Got it now?

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