# Solving z^n = C

• Apr 23rd 2009, 01:35 AM
Solving z^n = C
a) Express (z^5 - 1) as a product of two factors, one of which is linear

b) Find the zeroes of (z^5 - 1) giving your answers in the form r(cos *theta* + i sin *theta*) where r > 0 and -pi < *theta* < pi.

c) Express z^4 + z^3 + z^2 + z + 1 as a product of two real quadratic factors

• Apr 23rd 2009, 04:31 AM
Opalg
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a) Express (z^5 - 1) as a product of two factors, one of which is linear

z=1 is obviously a solution of $\displaystyle z^5-1=0$, so z–1 is a factor. Divide by it to find the other factor.

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b) Find the zeroes of (z^5 - 1) giving your answers in the form r(cos *theta* + i sin *theta*) where r > 0 and -pi < *theta* < pi.

The other four solutions of $\displaystyle z^5-1=0$ are the complex fifth roots of 1. So the values of θ will be 2kπ/5, for k=1,2,3,4.

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c) Express z^4 + z^3 + z^2 + z + 1 as a product of two real quadratic factors

This quartic polynomial should be the 'other factor" that you found in part a). So you know its zeros (from part b)) and therefore you know its (complex) linear factors. Pair these off into two complex conjugate pairs. The product of two complex conjugate complex linear factors will be a quadratic real factor.
• Apr 23rd 2009, 05:50 AM
Soroban

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a) Express $\displaystyle (z^5 - 1)$ as a product of two factors, one of which is linear.
$\displaystyle z^5 - 1 \;=\;(z-1)(z^4 + z^3 + z^2 + z + 1)$

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b) Find the zeroes of $\displaystyle (z^5 - 1)$ giving your answers in the form: .$\displaystyle r(\cos\theta + i\sin \theta)$
where: $\displaystyle r > 0$ and $\displaystyle -\pi < \theta < \pi$

We use DeMoivre's Theorem . . .

$\displaystyle z^5 \:=\:1 \quad\Rightarrow\quad z \:=\:1^{\frac{1}{5}}$ . . . We want the five $\displaystyle 5^{th}$ roots of 1.

They are given by: .$\displaystyle \cos\left(\tfrac{2\pi}{5}n\right) + i\sin\left(\tfrac{2\pi}{5}n\right)\,\text{ for }n = 0,1,2,3,4$

I'll let you write them out . . .

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c) Express $\displaystyle z^4 + z^3 + z^2 + z + 1$ as a product of two real quadratic factors.

We have: .$\displaystyle (z^2+az + 1)(z^2 + bz + 1) \:=\:z^4+z^3+z^2+z+1$

Then: .$\displaystyle z^4 + (a+b)z^3 + (ab+2)z^2 + (a+b)z + 1 \:=\:z^4+z^3+z^2+z+1$

Equate coefficients: .$\displaystyle \begin{Bmatrix}a + b &=& 1 \\ ab + 2 &=& 1 \end{Bmatrix}$

Solve the system: .$\displaystyle a \:=\:\frac{1+\sqrt{5}}{2},\quad b \:=\:\frac{1-\sqrt{5}}{2}$

Therefore: .$\displaystyle z^4+z^3+z^2+z+1 \;=\;\bigg[z^2 + \left(\tfrac{1+\sqrt{5}}{2}\right)z + 1\bigg]\,\bigg[x^2 + \left(\tfrac{1-\sqrt{5}}{2}\right)z + 1\bigg]$