Let $z = 1 + 2i$ and $w = 3 - i$. (a) (i) Find, in the form $x + iy$, $zw$. (ii) Find $\frac{10}{z}$. (b) Let $\alpha = 1 + i\sqrt{3}$ and $\beta = 1 + i$. (i) Find $\frac{\alpha}{\beta}$ in the form $x + iy$. (ii) Express $\alpha$ in modulus-argument form. (iii) Given that $\beta$ has the modulus-argument form $\beta = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$, find the modulus-argument form of $\frac{\alpha}{\beta}$. (iv) Hence find the exact value of $\sin \frac{\pi}{12}$. (c) Sketch the region in the complex plane where the inequalities $|z + w| \leq 1$ and $|z - i| \leq 1$ hold simultaneously. (d) The diagram shows two distinct points A and B that represent the complex numbers $z$ and $w$ respectively. The points A and B lie on the circle of radius $r$ centered at O. The point C representing the complex number $z + w$ also lies on this circle. (i) Using the fact that C lies on the circle, show geometrically that $\angle OAB = \frac{2\pi}{3}$. (ii) Hence show that $z^3 = w^3$. (iii) Show that $z^2 + w^2 + zw = 0$.

SSCE HSC Mathematics Extension 2 Complex numbers: Let $z = 1 + 2i$ and $w = 3

Let $z = 1 + 2i$ and $w = 3 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1

Question 2

Let $z = 1 + 2i$ and $w = 3 - i$. (a) (i) Find, in the form $x + iy$, $zw$. (ii) Find $\frac{10}{z}$. (b) Let $\alpha = 1 + i\sqrt{3}$ and $\beta = 1 + i$. (... show full transcript

Worked Solution & Example Answer:Let $z = 1 + 2i$ and $w = 3 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1

Step 1

Find, in the form $x + iy$, $zw$

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Answer

To find $zw$ , multiply the complex numbers:

zw = (1 + 2i)(3 - i) = 3 + 6i - i - 2 = 5 + 5i.

So, the answer is $5 + 5i$ .

Step 2

Find $\frac{10}{z}$

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To find $\frac{10}{z}$ , we compute:

\frac{10}{z} = \frac{10}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{10(1 - 2i)}{1 + 4} = \frac{10 - 20i}{5} = 2 - 4i.

Thus, $\frac{10}{z} = 2 - 4i.$

Step 3

Find $\frac{\alpha}{\beta}$ in the form $x + iy$

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For $\alpha = 1 + i\sqrt{3}$ and $\beta = 1 + i$ , we have:

\frac{\alpha}{\beta} = \frac{1 + i\sqrt{3}}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{(1 + i\sqrt{3})(1 - i)}{1 + 1} = \frac{1 - i + i\sqrt{3} + 1}{2} = \frac{2 + (\sqrt{3} - 1)i}{2} = 1 + \frac{\sqrt{3} - 1}{2}i.

Thus, $\frac{\alpha}{\beta} = 1 + \frac{\sqrt{3} - 1}{2}i$ .

Step 4

Express $\alpha$ in modulus-argument form

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To express $\alpha$ in modulus-argument form, calculate the modulus and argument:

|\alpha| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2,

and the argument is:

\arg(\alpha) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.

Thus, the modulus-argument form is:

$\alpha = 2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right).$

Step 5

Find the modulus-argument form of $\frac{\alpha}{\beta}$

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Using the result from part (iii), the modulus of $\beta$ is:

$|\beta| = \sqrt{1^2 + 1^2} = \sqrt{2},$

and the argument is:

Thus, we have:

$\frac{\alpha}{\beta} = \frac{2}{\sqrt{2}}\left(\cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{3} - \frac{\pi}{4}\right)\right) = \sqrt{2}\left(\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)\right).$

Step 6

Hence find the exact value of $\sin \frac{\pi}{12}$

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Using the angle subtraction formula, we have:

Step 7

Sketch the region in the complex plane where $|z + w| \leq 1$ and $|z - i| \leq 1$ hold simultaneously

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To sketch the regions, note that:

The inequality $|z + w| \leq 1$ describes a circle centered at the origin with radius 1.
The inequality $|z - i| \leq 1$ describes a circle centered at (0, 1) with radius 1.

The overlapping area of these two regions is the solution to both inequalities.

Step 8

Using the fact that C lies on the circle, show geometrically that $\angle OAB = \frac{2\pi}{3}$

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Since C lies on the circle, angles subtended by the same chord (AB) at the circumference are equal. Using properties of cyclic quadrilaterals, we have: $$\angle OAB = \angle OCB = \frac{1}{2} \cdot \text{arc } AB = \frac{2\pi}{3}.$

Step 9

Hence show that $z^3 = w^3$

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Since $\angle OAB = \frac{2\pi}{3}$ , it follows that: $$z^3 = |z|^3 \text{cis}(\angle OAB) = |w|^3 \text{cis}(\angle OAB) = w^3.$

Step 10

Show that $z^2 + w^2 + zw = 0$

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Using the identity: $z^2 + w^2 = (z + w)^2 - 2zw,$ we can rearrange to get: $z^2 + w^2 + zw = (z + w)^2 - 2zw + zw = (z + w)^2 - zw = 0.$

Let $z = 1 + 2i$ and $w = 3 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1

Worked Solution & Example Answer:Let $z = 1 + 2i$ and $w = 3 - i$ - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1

Find, in the form $x + iy$, $zw$

Find $\frac{10}{z}$

Find $\frac{\alpha}{\beta}$ in the form $x + iy$

Express $\alpha$ in modulus-argument form

Find the modulus-argument form of $\frac{\alpha}{\beta}$

Hence find the exact value of $\sin \frac{\pi}{12}$

Sketch the region in the complex plane where $|z + w| \leq 1$ and $|z - i| \leq 1$ hold simultaneously

Using the fact that C lies on the circle, show geometrically that $\angle OAB = \frac{2\pi}{3}$

Hence show that $z^3 = w^3$

Show that $z^2 + w^2 + zw = 0$

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