Complex numbers in polar

Hi all! Once again stuck on complex numbers...

(a) Expand (c+is)^4, where i^2 = -1. Hence by using De Moivre's theorem, show that

sin4@/cos4@ = (4c^3s - 4cs^3)/ (c^4 - 6c^2s^2 + s^4), where c = cos@, and s = sin @

I expanded to get c^4 + 4c^3si - 6c^2s^2 - 6cs^3i + s^4, and I see that sin4@/cos4@ is just s^4/c^4. But not sure how to show the solution.

(b) By substituting t = s/c = tan@, deduce an expression for tan4@ in terms of t only. Hence show that tan4@ = 0, when 4t - 4t^3 = 0

(c) Solve the equation obtained in part b for t. Hence find the values of @ in the range -pie/2 less than @ less than pie/2, for which tan4@ = 0.

Thanks once again for the help!

Hello, classicstrings!

You're that close in part (a) . . .

Quote:

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(a) Expand (cosθ + i·sinθ)^4, where i² = -1.
Hence by using De Moivre's theorem, show that:

. . sin4θ/cos4θ .= .(4c³s - 4cs³) / (c^4 - 6c²s² + s^4), where c = cosθ, and s = sinθ


I expanded to get: c^4 + 4c³s·i - 6c²s² - 6cs³·i + s^4,

and I see that: sin4θ/cos4θ is just s^4/c^4.
But not sure how to show the solution.

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DeMoivre's Theorem says: .(cosθ + i·sinθ)^4 .= .cos4θ + i·sin4θ

Your expansion is: .cos4θ + i·sin4θ .= .(c^4 - 6c²s² + s^4) + i·(4c³s - 4cs³)


Equate real and imaginary components:

. . cos4θ .= .c^4 - 6c²s² + s^4
. . sin4θ .= .4c³s - 4cs³

. . . . . . . . . sin4θ . . . . . . . 4c³s - 4cs³
Therefore: . -------- . = . ----------------------
. . . . . . . . . cos4θ . . . . c^4 - 6c²s² + s^4

Thanks for the help on the first part.

Still having problems with the next two parts. How do you simplify the expression, tanθ =

4c³s - 4cs³
----------------------
c^4 - 6c²s² + s^4

After substituting c = cosθ, s = sinθ

4c³s - 4cs³ = 4c³(s-1)
c^4 - 6c²s² + s^4 = (c-3cs)² - 9c²s² + s²

Hello again, classicstrings!

Quote:

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How do you simplify the expression:

. . . . . . . . . . . . . 4c³s - 4cs³
. . tan4θ . = . ----------------------
. . . . . . . . . .c^4 - 6c²s² + s^4

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Divide top and bottom by c^4:

. . . . . . . . . . . . .4c³s . . 4cs³
. . . . . . . . . . . . ------- - ------
. . . . . . . . . . . . .c^4 . . .c^4 . . . . . . . . . 4(s/c) - 4(s/c)³
. . tan4θ . = . ------------------------ . = . -------------------------
. . . . . . . . . . .c^4 . 6c²s² . . s^4 . . . . 1 - 6(s/c)² + (s/c)^4
. . . . . . . . . . .---- - ------- + -----
. . . . . . . . . . .c^4 . .c^4 . . .c^4


. . . . . . . . . . . . . . . . . . 4·tanθ - 4·tan³θ
Therefore: . tan4θ . = . -------------------------
. . . . . . . . . . . . . . . . .1 - 6·tanθ + tan^4θ

Thanks Soroban!

So the next part, of course when 4t - 4t^3 = 0, then the top is 0, and tan4θ = 0.

Now part c)

4t - 4t^3 = 4t(1-6t^2+t^4)
.
.
.
4t^3(t-root5)(t+root5)

Sub in t = tanθ,

So, θ = 0, t+-root5 = 0, θ = 65.9 degrees, - 65.9 degrees.

Would you please tell me if this is the right way Soroban? Thanks so much!

Hello again, classicstrings!

Quote:

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Solve the equation in part b for t.
Hence find the value of θ in the range -½π < θ < ½π for which tan4θ = 0.

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If tan4θ = 0, then: .4t - 4t³ .= .0

We have: .4·tanθ - 4·tan³θ .= .0

Factor: .4·tanθ (1 - tan²θ) .= .0


Solve the two equatons:

. . 4·tanθ = 0 . → . tanθ = 0 . → . θ = 0

. . 1 - tan²θ .= .0 . → . tan²θ = 1 . → . tanθ = ±1 . → . θ = ±¼π