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Parents Pricing Home SSCE HSC Mathematics Extension 2 The Argand diagram Let $ z = 1 + 2i $ and $ w = 3 - i
Let $ z = 1 + 2i $ and $ w = 3 - i - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1 Question 2
View full question Let $ z = 1 + 2i $ and $ w = 3 - i. $
Find, in the form $ x + iy, $
(i) $ zw $
(ii) $ rac{10}{z}.$
Let $ \alpha = 1 + i \sqrt{3} $ and $ \beta = 1 + i. $
(i) Fi... show full transcript
View marking scheme Worked Solution & Example Answer:Let $ z = 1 + 2i $ and $ w = 3 - i - HSC - SSCE Mathematics Extension 2 - Question 2 - 2004 - Paper 1
Find $ zw $ in the form $ x + iy $ Only available for registered users.
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To find z w zw z w
z w = ( 1 + 2 i ) ( 3 − i ) = 3 + 2 i − 1 − 6 i = 2 − 4 i . zw = (1 + 2i)(3 - i) = 3 + 2i - 1 - 6i = 2 - 4i. z w = ( 1 + 2 i ) ( 3 − i ) = 3 + 2 i − 1 − 6 i = 2 − 4 i . Thus, z w = 2 − 4 i zw = 2 - 4i z w = 2 − 4 i
Find $ \frac{10}{z} $ in the form $ x + iy $ Only available for registered users.
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To find 10 z \frac{10}{z} z 10
10 z = 10 1 + 2 i ⋅ 1 − 2 i 1 − 2 i = 10 ( 1 − 2 i ) 1 2 + ( 2 ) 2 = 10 ( 1 − 2 i ) 5 = 2 − 4 i . \frac{10}{z} = \frac{10}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{10(1 - 2i)}{1^2 + (2)^2} = \frac{10(1 - 2i)}{5} = 2 - 4i. z 10 = 1 + 2 i 10 ⋅ 1 − 2 i 1 − 2 i = 1 2 + ( 2 ) 2 10 ( 1 − 2 i ) = 5 10 ( 1 − 2 i ) = 2 − 4 i . So, 10 z = 2 − 4 i \frac{10}{z} = 2 - 4i z 10 = 2 − 4 i
Find $ \frac{\alpha}{\beta} $ in the form $ x + iy $ Only available for registered users.
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We have:
α = 1 + i 3 , β = 1 + i . α β = ( 1 + i 3 ) ( 1 − i ) ( 1 + i ) ( 1 − i ) = ( 1 − i + i 3 + 1 3 ) 2 = 2 + ( 3 − 1 ) i 2 = 1 + ( 3 − 1 ) 2 i . \alpha = 1 + i\sqrt{3}, \, \beta = 1 + i. \
\frac{\alpha}{\beta} = \frac{(1 + i\sqrt{3})(1 - i)}{(1 + i)(1 - i)} = \frac{(1 - i + i\sqrt{3} + 1\sqrt{3})}{2} = \frac{2 + (\sqrt{3} - 1)i}{2} = 1 + \frac{(\sqrt{3} - 1)}{2}i. α = 1 + i 3 , β = 1 + i . β α = ( 1 + i ) ( 1 − i ) ( 1 + i 3 ) ( 1 − i ) = 2 ( 1 − i + i 3 + 1 3 ) = 2 2 + ( 3 − 1 ) i = 1 + 2 ( 3 − 1 ) i .
Express $ \alpha $ in modulus-argument form Only available for registered users.
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To determine the modulus and argument of α \alpha α
Modulus:
∣ α ∣ = 1 2 + ( 3 ) 2 = 2. \lvert \alpha \rvert = \sqrt{1^2 + (\sqrt{3})^2} = 2. ∣ α ∣ = 1 2 + ( 3 ) 2 = 2.
Argument:
θ = tan − 1 ( 3 1 ) = π 3 . \theta = \tan^{-1}\left( \frac{\sqrt{3}}{1} \right) = \frac{\pi}{3}. θ = tan − 1 ( 1 3 ) = 3 π .
Thus, α = 2 ( cos π 3 + i sin π 3 ) . \alpha = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right). α = 2 ( cos 3 π + i sin 3 π ) .
Given $ \beta $ find the modulus-argument form of $ \frac{\alpha}{\beta} $ Only available for registered users.
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We already have α β \frac{\alpha}{\beta} β α β = 2 ( cos π 4 + i sin π 4 ) : \beta = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right): β = 2 ( cos 4 π + i sin 4 π ) :
α β = 2 2 ( cos ( π 3 − π 4 ) + i sin ( π 3 − π 4 ) ) = 2 ( cos π 12 + i sin π 12 ) . \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 \frac{\pi}{12} + i \sin \frac{\pi}{12} \right). β α = 2 2 ( cos ( 3 π − 4 π ) + i sin ( 3 π − 4 π ) ) = 2 ( cos 12 π + i sin 12 π ) .
Hence find the exact value of $ \sin \frac{\pi}{12} $ Only available for registered users.
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From the modulus-argument form, we can derive:
sin π 12 = 1 − 3 2 . \sin \frac{\pi}{12} = \frac{1 - \sqrt{3}}{2}. sin 12 π = 2 1 − 3 .
Sketch the region where $ |z + w| \leq 1 $ and $ |z - i| \leq 1 $ hold simultaneously Only available for registered users.
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The region can be sketched by taking the intersection of two circles:
∣ z + w ∣ ≤ 1 |z + w| \leq 1 ∣ z + w ∣ ≤ 1 − w -w − w 1 1 1 ∣ z − i ∣ ≤ 1 |z - i| \leq 1 ∣ z − i ∣ ≤ 1 i i i 1 1 1
The overlap of these circles indicates the solution region.
Using the fact that C lies on the circle, show geometrically that $ \angle OAC = \frac{2\pi}{3} $ Only available for registered users.
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Using properties of a circle and isosceles triangles, we can show that:
The angle at center O is twice the angle at the circumference. This can be represented as:
∠ O A C = 2 ⋅ ∠ A B C . \angle OAC = 2 \cdot \angle ABC. ∠ O A C = 2 ⋅ ∠ A BC .
Thus, since B B B C C C ∠ O A C = 2 π 3 . \angle OAC = \frac{2\pi}{3}. ∠ O A C = 3 2 π .
Hence show that $ z^3 = w^3 $ Only available for registered users.
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Using the established angles and the symmetry of the points:
∠ O A B = ∠ O B C . \angle OAB = \angle OBC. ∠ O A B = ∠ OBC . A A A B B B
Hence, z 3 = w 3 z^3 = w^3 z 3 = w 3
Show that $ z^2 + w^2 + zw = 0 $ Only available for registered users.
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Given z 3 = w 3 z^3 = w^3 z 3 = w 3 z 3 − w 3 = ( z − w ) ( z 2 + z w + w 2 ) = 0. z^3 - w^3 = (z - w)(z^2 + zw + w^2) = 0. z 3 − w 3 = ( z − w ) ( z 2 + z w + w 2 ) = 0. . . .
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