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) Find, in the form $x + iy$,
(i) $zw$.
(ii) $\frac{10}{z}$.
(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 $x + iy$, (i) $zw$
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Answer
To find zw, we multiply the two complex numbers:
zw=(1+2i)(3−i)=3+6i−1i−2i2=3+6i+2=5+5i.
Thus, the answer is 5+5i.
Step 2
Find, in the form $x + iy$, (ii) $\frac{10}{z}$
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Answer
To compute z10, we need to multiply the numerator and the denominator by the conjugate of z:
Let $\alpha = 1 + i\sqrt{3}$ and $\beta = 1 + i$. (ii) Express $\alpha$ in modulus-argument form.
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Answer
First, we find the modulus of α:
∣α∣=(1)2+(3)2=1+3=2.
Next, the argument of α is given by:
θ=tan−1(13)=3π.
Thus, α in modulus-argument form is:
α=2(cos3π+isin3π).
Step 5
Let $\beta$ has the modulus-argument form (iii) Given that $\beta = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$, find the modulus-argument form of $\frac{\alpha}{\beta}$.
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Answer
To find the modulus-argument form of βα:
The modulus:
∣β∣∣α∣=22=2.
The argument:
arg(α)−arg(β)=3π−4π=124π−3π=12π.
Therefore:
βα=2(cos12π+isin12π).
Step 6
Hence find the exact value of $\sin\frac{\pi}{12}$. (iv)
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Answer
Using the sine function from the modulus-argument form;
$$\sin\frac{\pi}{12} = \sin(\frac{\pi}{3} - \frac{\pi}{4}) = \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}.$
Step 7
Sketch the region in the complex plane where the inequalities (c) $|z + z| \leq 1$ and $|z - i| \leq 1$ hold simultaneously.
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Answer
The inequality ∣z+z∣≤1 defines the area inside the circle centered at the origin with radius 1.
The inequality ∣z−i∣≤1 defines the area inside the circle centered at (0, 1) with radius 1.
The intersection of these two regions represents the solution set, which is a lens-shaped area in the complex plane.
Step 8
Using the fact that C lies on the circle, show geometrically that (d) $\angle OAC = \frac{2\pi}{3}$.
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Since point C lies on the circle, the segments OA and OC must meet at right angles.
The angles AOB and AOC form a triangle with angle AOC being the external angle:
Thus:
∠OAC+∠AOB+∠AOC=(32π+2∠OAC=π⇒∠OAC=32π.
Step 9
Hence show that (ii) $z^3 = w^3$.
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Answer
From above:
Using the geometric relationships on angles:
If ∠OAC=32π, hence it follows that:
where both have same r, leading us to conclude:
z3=w3.
Step 10
Show that (iii) $z^2 + w^2 + 2zw = 0$.
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From the concept of complex multiplication of angles and magnitudes, we have:
(z+w)2=z2+w2+2zw=0
This proves that the expression sums to yield zero.
