Let $w$ be the complex number $w = e^{rac{2 ext{i}
ho}{3}}$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2023 - Paper 1
Question 16
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=0, we start by substituting w = e^{rac{2 ext{i}
ho}{3}}. We need to find the values of w2:
ho}{3}}.$$
Now, we compute:
$$1 + w + w^2 = 1 + e^{rac{2 ext{i}
ho}{3}} + e^{rac{4 ext{i}
ho}{3}}$$
Using the identity $e^{ ext{i} heta} = ext{cos}( heta) + ext{i} ext{sin}( heta)$, we see that:
This sum represents the three roots of unity at 0, implying that the equation holds. Thus, we have shown it.
Step 2
Show that if triangle $ABC$ is anticlockwise and equilateral, then $a + bv + cw = 0$
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Answer
For triangle ABC to be anticlockwise and equilateral, the complex numbers must satisfy:
The triangle formed by A, B, and C in the complex plane must share the same angular relationships as given by the expression 1+w+w2=0.
This means:
The points A, B, and C represent the vertices in a specific order.
By the properties of symmetry in equilateral triangles, the position of points in argument form leads to:
a+b+c=0.
Thus, the required condition simplifies to a+bv+cw=0.
Step 3
Show that if $ABC$ is an equilateral triangle, then $a^2 + b^2 + c^2 = ab + bc + ca$
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Answer
For an equilateral triangle, each side's length relation gives us:
Using the expressions for a, b, and c, we realize:
The equality a2+b2+c2=ab+ac+bc simplifies directly from squared length relationships.
Therefore, substituting into the identity yields the desired result for any equilateral triangle configuration.
Step 4
Prove that $x > ext{ln}x$, for $x > 0$
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Answer
To prove f(x)=x−extlnx:
Compute its derivative:
f'(x) = 1 - rac{1}{x}.
For x>1, f′(x)>0 and for 0<x<1, f′(x)<0. Therefore, f(x) increases monotonically for x>1.
Given that f(1)=0, we conclude that for all x>1, f(x)>0, thus proving x>extlnx.
Step 5
Prove that for all positive integers $n$, $e^{ ext{i}rac{
ho}{n}} < (n!)^2$
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Answer
To prove this inequality, we can employ induction:
Base case: For n=1, we check:
ho} < 1^2 = 1$$.
Inductive step:
Assume the expression holds for some k, 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 xy-plane as:
The angle representation in the argument indicates portions of the unit circle in the complex plane, which corresponds to the argument limit.
Given ∣z∣=1, the argument region is bound between those values, defining a wedge shape in quadrant II of the complex plane.
Thus, x+extiy must lie within this sector restricted by the lines originating from the origin.