3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$. Find the value of $p$ and the value of $q$. (b) $v = 2 - 2\sqrt{3}i$. Write $v$ in the form $r(cos \theta + i sin \theta)$, where $r \in \mathbb{R}$ and $0 \leq \theta \leq 2\pi$. (ii) Use your answer to part (b)(i) to find the two possible values of $w$, where $w^2 = v$. Give your answers in the form $a + ib$, where $a, b \in \mathbb{R}$.

Leaving Cert Mathematics Complex Numbers: 3 + 2i is a root of $z^2 + pz +

3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$ - Leaving Cert Mathematics - Question 5 - 2019

Question 5

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3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$. Find the value of $p$ and the value of $q$. (b) $v = 2 - 2\sqrt{3}i$. Write $... show full transcript

Worked Solution & Example Answer:3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$ - Leaving Cert Mathematics - Question 5 - 2019

Step 1

Find the values of $p$ and $q$

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Answer

To find the values of $p$ and $q$ , we first note that if $3 + 2i$ is a root, its conjugate $3 - 2i$ is also a root. Using Vieta's formulas, we have:

The sum of the roots is given by $3 + 2i + (3 - 2i) = 6 \Rightarrow -p = 6 \Rightarrow p = -6$
The product of the roots is given by $(3 + 2i)(3 - 2i) = 9 + 4 = 13 \Rightarrow q = 13$

Step 2

Write $v$ in the form $r(cos \theta + i sin \theta)$

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Answer

To express $v = 2 - 2\sqrt{3}i$ in polar form:

Calculate the modulus: $r = |v| = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$
Calculate the argument: $\theta = \arctan\left(\frac{-2\sqrt{3}}{2}\right) = \arctan(-\sqrt{3}) = 300^\circ = \frac{5\pi}{3}$

Thus, ( v = 4 \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) )

Step 3

Find the two possible values of $w$

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Answer

Since $w^2 = v$ , we apply De Moivre's theorem: $w = \sqrt{4} \left( \cos\left(\frac{5\pi}{3} + k\pi\right) + i \sin\left(\frac{5\pi}{3} + k\pi\right) \right) \text{ for } k = 0, 1$

For $k = 0$ : $w = 2 \left( \cos\frac{5\pi}{3} + i \sin\frac{5\pi}{3} \right) = 2 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 1 - \sqrt{3} i$

For $k = 1$ : $w = 2 \left( \cos\left(\frac{5\pi}{3} + \pi\right) + i \sin\left(\frac{5\pi}{3} + \pi\right) \right) = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -1 + \sqrt{3} i$

Thus, the two possible values of $w$ are:

$1 - \sqrt{3} i$
$-1 + \sqrt{3} i$

3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$ - Leaving Cert Mathematics - Question 5 - 2019

Worked Solution & Example Answer:3 + 2i is a root of $z^2 + pz + q = 0$, where $p, q \in \mathbb{R}$, and $i^2 = -1$ - Leaving Cert Mathematics - Question 5 - 2019

Find the values of $p$ and $q$

Write $v$ in the form $r(cos \theta + i sin \theta)$

Find the two possible values of $w$

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