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Parents Pricing Home SSCE HSC Mathematics Extension 2 3D space Consider the complex numbers $z = -2 - 2i$ and $w = 3 + i$
Consider the complex numbers $z = -2 - 2i$ and $w = 3 + i$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2014 - Paper 1 Question 11
View full question Consider the complex numbers $z = -2 - 2i$ and $w = 3 + i$.
(i) Express $z + w$ in modulus–argument form.
(ii) Express $ rac{z}{w}$ in the form $x + iy$, where $x$... show full transcript
View marking scheme Worked Solution & Example Answer:Consider the complex numbers $z = -2 - 2i$ and $w = 3 + i$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2014 - Paper 1
Express $z + w$ in modulus–argument form Only available for registered users.
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To express z + w z + w z + w
Calculate z + w z + w z + w z + w = ( − 2 − 2 i ) + ( 3 + i ) = 1 − i . z + w = (-2 - 2i) + (3 + i) = 1 - i. z + w = ( − 2 − 2 i ) + ( 3 + i ) = 1 − i .
Find the modulus:∣ z + w ∣ = 1 2 + ( − 1 ) 2 = 2 . |z + w| = \sqrt{1^2 + (-1)^2} = \sqrt{2}. ∣ z + w ∣ = 1 2 + ( − 1 ) 2 = 2 .
Find the argument:arg ( z + w ) = tan − 1 ( − 1 1 ) = − π 4 . \arg(z + w) = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}. arg ( z + w ) = tan − 1 ( 1 − 1 ) = − 4 π .
Thus, the modulus–argument form is:
Express $\frac{z}{w}$ in the form $x + iy$ Only available for registered users.
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To find z w \frac{z}{w} w z
Compute:
− 2 − 2 i 3 + i = ( − 2 − 2 i ) ( 3 − i ) ( 3 + i ) ( 3 − i ) = − 6 + 2 − 6 i − 2 i 9 + 1 = − 4 − 8 i 10 . \frac{-2 - 2i}{3 + i} = \frac{(-2 - 2i)(3 - i)}{(3 + i)(3 - i)} = \frac{-6 + 2 - 6i - 2i}{9 + 1} = \frac{-4 - 8i}{10}. 3 + i − 2 − 2 i = ( 3 + i ) ( 3 − i ) ( − 2 − 2 i ) ( 3 − i ) = 9 + 1 − 6 + 2 − 6 i − 2 i = 10 − 4 − 8 i .
Simplify:
z w = − 0.4 − 0.8 i . \frac{z}{w} = -0.4 - 0.8i. w z = − 0.4 − 0.8 i .
Thus, in the form x + i y x + iy x + i y − 0.4 − 0.8 i . -0.4 - 0.8i. − 0.4 − 0.8 i .
Evaluate $\int_{0}^{1} (3x - 1)\cos(\pi x) \, dx$ Only available for registered users.
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To evaluate the integral:
We can use integration by parts where:
Let:
u = ( 3 x − 1 ) ⇒ d u = 3 d x , u = (3x - 1) \, \Rightarrow \, du = 3 \, dx, u = ( 3 x − 1 ) ⇒ d u = 3 d x , d v = cos ( π x ) d x ⇒ v = 1 π sin ( π x ) . dv = \cos(\pi x) \, dx \Rightarrow \, v = \frac{1}{\pi}\sin(\pi x). d v = cos ( π x ) d x ⇒ v = π 1 sin ( π x ) .
Thus:
∫ u d v = u v − ∫ v d u \int u \, dv = uv - \int v \, du ∫ u d v = uv − ∫ v d u
Then substituting:
= ( 3 x − 1 ) 1 π sin ( π x ) ∣ 0 1 − ∫ sin ( π x ) ⋅ 3 d x = (3x-1) \frac{1}{\pi} \sin(\pi x) \bigg|_{0}^{1} - \int \sin(\pi x) \cdot 3 \, dx = ( 3 x − 1 ) π 1 sin ( π x ) 0 1 − ∫ sin ( π x ) ⋅ 3 d x
After solving the above and integrating:
At x = 1 x = 1 x = 1 0 0 0 s i n ( π ) = 0 sin(\pi) = 0 s in ( π ) = 0
At x = 0 x = 0 x = 0 0 0 0 s i n ( 0 ) = 0 sin(0) = 0 s in ( 0 ) = 0
Therefore:
= − 3 π [ − cos ( π x ) ] ∣ 0 1 → − 3 π [ − ( − 1 − 1 ) ] = 6 π . = -\frac{3}{\pi}[-\cos(\pi x)]\bigg|_0^1\rightarrow -\frac{3}{\pi}[-(-1-1)] = \frac{6}{\pi}. = − π 3 [ − cos ( π x )] 0 1 → − π 3 [ − ( − 1 − 1 )] = π 6 .
Sketch the region in the Argand diagram where $|z| \leq |2 - 2i|$ and $\frac{\pi}{4} \leq \arg z \leq \frac{\pi}{4}$ Only available for registered users.
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To sketch the region:
Calculate the modulus:
∣ 2 − 2 i ∣ = 2 2 + ( − 2 ) 2 = 8 = 2 2 . |2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}. ∣2 − 2 i ∣ = 2 2 + ( − 2 ) 2 = 8 = 2 2 .
The inequality ∣ z ∣ ≤ 2 2 |z| \leq 2\sqrt{2} ∣ z ∣ ≤ 2 2 2 2 2\sqrt{2} 2 2
The arguments indicate that z z z π 4 \frac{\pi}{4} 4 π π 4 \frac{\pi}{4} 4 π
Sketching both will show a circular sector from the origin out to radius 2 2 2\sqrt{2} 2 2
Without the use of calculus, sketch the graph $y = \frac{x^2 - 1}{x^2}$ showing all intercepts Only available for registered users.
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To sketch the graph:
Identify intercepts:
X-intercepts occur when y = 0 y = 0 y = 0 x 2 − 1 x 2 = 0 ⇒ x 2 − 1 = 0 ⇒ x = ± 1. \frac{x^2 - 1}{x^2} = 0 \Rightarrow x^2 - 1 = 0 \Rightarrow x = \pm 1. x 2 x 2 − 1 = 0 ⇒ x 2 − 1 = 0 ⇒ x = ± 1. Y-intercept occurs at x = 0 x = 0 x = 0 y = 0 − 1 0 (undefined) y = \frac{0 - 1}{0} \text{ (undefined)} y = 0 0 − 1 (undefined)
Asymptote Analysis:
As x → ± ∞ x \rightarrow \pm \infty x → ± ∞ y → 1. y \rightarrow 1. y → 1.
Vertical asymptote at x = 0 x = 0 x = 0 0 0 0
Sketch the graph noting the intercepts and asymptotic behavior.
Using the method of cylindrical shells, or otherwise, find the volume of the solid Only available for registered users.
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To find the volume:
The region bounded by the curve y = 6 − y y = 6 - y y = 6 − y y y y y = 3 y = 3 y = 3
The volume of revolution using cylindrical shells is given by:
V = 2 π ∫ a b x ( height ) d x V = 2\pi \int_{a}^{b} x(\text{height}) \, dx V = 2 π ∫ a b x ( height ) d x h = 6 − y h = 6 - y h = 6 − y
Integrate:
Set up the integral, integrating bounds are determined by intercepts.
Calculate to find the volume of the solid.
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