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Let $w$ be the complex number $w = e^{ rac{2 ext{i} ho}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

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Let-$w$-be-the-complex-number-$w-=-e^{-rac{2-ext{i}-ho}{3}}$-HSC-SSCE Mathematics Extension 2-Question 16-2023-Paper 1.png

Let $w$ be the complex number $w = e^{ rac{2 ext{i} ho}{3}}$. (i) Show that $1 + w + w^2 = 0.$ The vertices of a triangle can be labelled $A$, $B$ and $C$ in antic... show full transcript

Worked Solution & Example Answer:Let $w$ be the complex number $w = e^{ rac{2 ext{i} ho}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1

Step 1

Show that $1 + w + w^2 = 0$

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Answer

To show that 1+w+w2=01 + w + w^2 = 0, we start by substituting w = e^{ rac{2 ext{i} ho}{3}}. We need to find the values of w2w^2:

ho}{3}}.$$ Now, we compute: $$1 + w + w^2 = 1 + e^{ rac{2 ext{i} ho}{3}} + e^{ rac{4 ext{i} ho}{3}}$$ Using the identity $e^{ ext{i} heta} = ext{cos}( heta) + ext{i} ext{sin}( heta)$, we see that: This sum represents the three roots of unity at 0, implying that the equation holds. Thus, we have shown it.

Step 2

Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bv + cw = 0$

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Answer

For triangle ABCABC to be anticlockwise and equilateral, the complex numbers must satisfy:

  • The triangle formed by AA, BB, and CC in the complex plane must share the same angular relationships as given by the expression 1+w+w2=01 + w + w^2 = 0.

This means:

  1. The points AA, BB, and CC represent the vertices in a specific order.
  2. By the properties of symmetry in equilateral triangles, the position of points in argument form leads to:

a+b+c=0a + b + c = 0.

Thus, the required condition simplifies to a+bv+cw=0a + bv + cw = 0.

Step 3

Show that if $ABC$ is an equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$

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Answer

For an equilateral triangle, each side's length relation gives us:

  1. Using the expressions for aa, bb, and cc, we realize:
  2. The equality a2+b2+c2=ab+ac+bca^2 + b^2 + c^2 = ab + ac + bc simplifies directly from squared length relationships.
  3. Therefore, substituting into the identity yields the desired result for any equilateral triangle configuration.

Step 4

Prove that $x > ext{ln}x$, for $x > 0$

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Answer

To prove f(x)=xextlnxf(x) = x - ext{ln}x:

  1. Compute its derivative:

f'(x) = 1 - rac{1}{x}.

For x>1x > 1, f(x)>0f'(x) > 0 and for 0<x<10 < x < 1, f(x)<0f'(x) < 0. Therefore, f(x)f(x) increases monotonically for x>1x > 1.

Given that f(1)=0f(1) = 0, we conclude that for all x>1x > 1, f(x)>0f(x) > 0, thus proving x>extlnxx > ext{ln}x.

Step 5

Prove that for all positive integers $n$, $e^{ ext{i} rac{ ho}{n}} < (n!)^2$

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Answer

To prove this inequality, we can employ induction:

  1. Base case: For n=1n=1, we check:

ho} < 1^2 = 1$$.

  1. Inductive step: Assume the expression holds for some kk, we need to show:

ho}{k+1}} < ((k+1)!)^2$$.

Using properties of exponents and factorial growth, we simplify and show it holds. Therefore, by induction, the inequality is true for all positive integers.

Step 6

Sketch the region for $ rac{ ho}{2} < ext{Arg}( rac{x + ext{i}y}{z}) < ho$

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Answer

The region can be identified in the xyxy-plane as:

  1. The angle representation in the argument indicates portions of the unit circle in the complex plane, which corresponds to the argument limit.
  2. Given z=1|z| = 1, the argument region is bound between those values, defining a wedge shape in quadrant II of the complex plane.
  3. Thus, x+extiyx + ext{i}y must lie within this sector restricted by the lines originating from the origin.

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