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Let $z = 2 - i ext{√}3$ and $w = 1 + i ext{√}3$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2013 - Paper 1

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Question 11

Let-$z-=-2---i-ext{√}3$-and-$w-=-1-+-i-ext{√}3$-HSC-SSCE Mathematics Extension 2-Question 11-2013-Paper 1.png

Let $z = 2 - i ext{√}3$ and $w = 1 + i ext{√}3$. (i) Find $z + w$. (ii) Express $w$ in modulus-argument form. (iii) Write $w^{24}$ in its simplest form. (b) Find... show full transcript

Worked Solution & Example Answer:Let $z = 2 - i ext{√}3$ and $w = 1 + i ext{√}3$ - HSC - SSCE Mathematics Extension 2 - Question 11 - 2013 - Paper 1

Step 1

Find $z + w$

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Answer

To find z+wz + w, we compute:

z+w=(2i3)+(1+i3)=2+1+(i3+i3)=3.z + w = (2 - i\sqrt{3}) + (1 + i\sqrt{3}) = 2 + 1 + (-i\sqrt{3} + i\sqrt{3}) = 3.

Thus, the answer is 33.

Step 2

Express $w$ in modulus-argument form

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Answer

To express ww in modulus-argument form, we calculate its modulus:

w=12+(3)2=1+3=4=2.|w| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2.

Next, we find the argument:

arg(w)=tan1(31)=π3.\arg(w) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}.

Therefore, the modulus-argument form of ww is:

w=2cis(π3).w = 2\text{cis}\left(\frac{\pi}{3}\right).

Step 3

Write $w^{24}$ in its simplest form

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Answer

From the modulus-argument form, we can find:

w24=(2cis(π3))24=224cis(24π3)=16777216cis(8π).w^{24} = \left(2 \text{cis}\left(\frac{\pi}{3}\right)\right)^{24} = 2^{24} \text{cis}(24 \cdot \frac{\pi}{3}) = 16777216 \text{cis}(8\pi).

Since extcis(8π)=1 ext{cis}(8\pi) = 1, the simplest form of w24w^{24} is:

w24=16777216.w^{24} = 16777216.

Step 4

Find numbers $A$, $B$ and $C$ such that...

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Answer

To find AA, BB, and CC, we can write:

x2+8x+11=A(x2+2)+(Bx+C)(x3).x^2 + 8x + 11 = A(x^2 + 2) + (Bx + C)(x - 3).

Expanding the right-hand side gives:

Ax2+2A+Bx23Bx+Cx3C=(A+B)x2+(C3B)x+(2A3C).Ax^2 + 2A + Bx^2 - 3Bx + Cx - 3C = (A + B)x^2 + (C - 3B)x + (2A - 3C).

By equating coefficients:

  1. A+B=1A + B = 1
  2. C3B=8C - 3B = 8
  3. 2A3C=112A - 3C = 11

Solving this system, we find:

A=4,B=3,C=1.A = 4, B = -3, C = -1.

Step 5

Factorise $z^2 + 4iz + 5$

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Answer

To factorise z2+4iz+5z^2 + 4iz + 5, we use the quadratic formula:

z=b±b24ac2a,z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

where a=1a = 1, b=4ib = 4i, and c=5c = 5:

  1. Calculate the discriminant: b24ac=(4i)220=1620=36.b^2 - 4ac = (4i)^2 - 20 = -16 - 20 = -36.
  2. Compute the roots:

z=4i±362=2i±3.z = \frac{-4i \pm \sqrt{-36}}{2} = 2i \pm 3.

Thus, the factorization is:

(z(3+2i))(z(32i)).(z - (3 + 2i))(z - (3 - 2i)).

Step 6

Evaluate $\int_0^1 x^3 \sqrt{1 - x^2} \, dx$

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Answer

To evaluate this integral, we use the substitution x=sinθx = \sin \theta, then dx=cosθdθdx = \cos \theta \, d\theta.

0 to 1 maps to 0 to π2\frac{\pi}{2}:

0π2sin3θ1sin2θcosθdθ=0π2sin3θcos2θdθ.\int_0^{\frac{\pi}{2}} \sin^3 \theta \sqrt{1 - \sin^2 \theta} \cos \theta \, d\theta = \int_0^{\frac{\pi}{2}} \sin^3 \theta \cos^2 \theta \, d\theta.

Using the reduction formula or direct integration:

The value of the integral is:

38.\frac{3}{8}.

Step 7

Sketch the region on the Argand diagram defined by $z^2 + z^2 \leq 8$

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Answer

The given inequality can be rewritten as:

2z28    z24.2z^2 \leq 8 \quad \implies \quad z^2 \leq 4.

This inequality represents a circle of radius 2 centered at the origin in the Argand diagram. Therefore, the region is the area inside and on the circle defined by:

z2.|z| \leq 2.

The sketch should represent a filled circle with radius 2.

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