smc-1050-10-Quadratic roots

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that \(1+w+w^2+\cdots+w^6=0\).

Show that \(w\) is a 7th root of unity. (1 mark)

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The complex number \(\alpha=w+w^2+w^4\) is a root of the equation \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

Find the other root of \(x^2+b x+c=0\) in terms of positive powers of \(w\). (2 marks)

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Find the numerical value of \(c\). (1 mark)

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i. \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\)	\(=0\)
\(1-w^7\)	\(=0\)
\(w^7=1\)	\(=1\)

ii. \(w^6+w^5+w^3\)

iii. \(2\)

Show Worked Solution

i. \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\)	\(=0\)
\(1-w^7\)	\(=0\)
\(w^7\)	\(=1\)

ii. \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\)	\(=\overline{w+w^2+w^4}\)
	\(=\overline{w}+\overline{w^2}+\overline{w^4}\)
	\(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
	\(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
	\(=w^6+w^5+w^3\)

iii. \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\)	\(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
	\(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
	\(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
	\(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
	\(=2\)

Complex Numbers, EXT2 N2 2022 SPEC1 1

Consider the equation `p(z)=z^2 + 6iz-25`, `z ∈ C`.

Express `p(z)` in the form `p(z) = (z+ai)^2 + b` where `a`, ` b ∈ R`. (1 mark)
Hence, or otherwise, find the solutions of the equation `p(z) = 0`. (2 marks)

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`p(z) = (z + 3i)^2-16`
`z = ± 4-3i`

Show Worked Solution

a.	`p(z)`	`= (z + 3i)^2-(3i)^2-25`
		`=(z + 3i)^2 +9-25`
		`=(z + 3i)^2-16`

b.	`(z + 3i)^2`	`= 16`
	`z + 3i`	`= ± 4`
	`z `	`= ± 4-3i`

Complex Numbers, EXT2 N2 2023 HSC 11a

Solve the quadratic equation

\(z^2-3 z+4=0\)

where \(z\) is a complex number. Give your answers in Cartesian form. (2 marks)

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\(\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)

Show Worked Solution

\(z^2-3 z+4=0\)

\(z\)	\(=\dfrac{3 \pm \sqrt{(-3)^2-4 \times 1 \times 4}}{2} \)
	\(=\dfrac{3 \pm \sqrt{-7}}{2} \)
	\(=\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)

Complex Numbers, EXT2 N2 2022 HSC 13c

Consider the equation `z^5+1=0`, where `z` is a complex number.

Solve the equation `z^5+1=0` by finding the 5th roots of `-1`. (2 marks)

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Show that if `z` is a solution of `z^5+1=0` and `z !=-1`, then `u=z+(1)/(z)` is a solution of `u^2-u-1=0`. (2 marks)

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Hence find the exact value of `cos\ (3pi)/(5)`. (3 marks)

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`z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
`text{Proof (See Worked Solutions)}`
`(1-sqrt5)/4`

Show Worked Solution

i. `z^5+1=0\ \ =>\ \ z^5=-1`

`z=e^(i((2k+1)/5)),\ \ kin{0,1,-1,2,-2}`

`:.z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`

ii. `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`

`text{Divide by}\ z^2\ \ (z!=0)`

`z^2-z+1-1/z+1/z^2`	`=0`
`z^2+1/z^2-(z+1/z)+1`	`=0`
`z^2+2+1/z^2-(z+1/z)-1`	`=0`
`(z+1/z)^2-(z-1/z)-1`	`=0`

`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`

Mean mark (ii) 53%.

iii. `u^2-u-1=0`

`text{By quadratic formula:}`

`u`	`=(1+-sqrt(1-4xx1xx(-1)))/(2)`
	`=(1+-sqrt5)/2`

♦ Mean mark (iii) 43%.

`z+1/z`	`=(1+-sqrt5)/2`
`e^(i(3pi)/5)+e^(-i(3pi)/5)`	`=(1-sqrt5)/2,\ \ (cos\ (3pi)/5 <0)`
`2cos((3pi)/5)`	`=(1-sqrt5)/2`
`cos((3pi)/5)`	`=(1-sqrt5)/4`

Complex Numbers, EXT2 N2 2021 SPEC1 8

Solve `z^2 + 2z + 2 = 0` for `z`, where `z ∈ C`. (1 mark)

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Solve `z^2 + 2barz + 2 = 0` for `z`, where `z ∈ C`. (3 marks)

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`z = -1 – i\ \ text(or)\ \ -1 + i`
`z = 1 ± sqrt5 i`

Show Worked Solution

a.	`z^2 + 2z + 2`	`= 0`
	`z^2 + 2z + 1 + 1`	`= 0`
	`(z + 1)^2 + 1`	`= 0`
	`(z + 1)^2 – i^2`	`= 0`
	`(z + 1 + i)(z + 1 – i)`	`= 0`

`:. z = -1 – i\ \ \ text(or)\ \ -1 + i`

b. `z = x + yi \ => \ barz = x – yi`

`z^2 + 2barz + 2`	`= 0`
`(x + yi)^2 + 2(x – yi) + 2`	`= 0`
`x^2 + 2xyi – y^2 + 2x – 2yi + 2`	`= 0`
`x^2 – y^2 + 2x + 2 + (2xy – 2y)i`	`= 0`

`text(Find)\ \ x, y\ text(such that)`

`x^2 – y^2 + 2x + 2`	`= 0\ …\ (1)`
`2xy – 2y`	`= 0\ …\ (2)`

`text(When)\ \ 2xy – 2y = 0`

`2y(x – 1)`	`= 0`
`x`	`= 1`

`text(Substitute)\ \ x = 1\ \ text{into (1)}`

`1 – y^2 + 2 + 2`	`= 0`
`y^2`	`= 5`
`y`	`= ±sqrt5`

`:. z = 1 ± sqrt5 i`

Complex Numbers, EXT2 N2 2021 HSC 11d

Find the two square roots of `−i` , giving the answers in the form `x + iy`, where `x` and `y` are real numbers. (2 marks)

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Hence, or otherwise, solve `z^2 + 2z + 1 + i = 0` giving your solutions in the form `a + i b` where `a` and `b` are real numbers. (2 marks)

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`z = 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = – 1/sqrt2+ 1/sqrt2 i`
`z = -1 + 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = -1 – 1/sqrt2+ 1/sqrt2 i`

Show Worked Solution

i. `text{Solution 1}`

`z = x + iy \ , \ z^2 = -i`

`z^2 = x^2 – y^2 + 2x yi = -i`

`x^2 – y^2 = 0 \ …\ (1)`

`2xy = -1 \ …\ (2)`

`x= ± y \ …\ (1)′`

`text{Substitute} \ \ x = y \ \ text{into} \ (2)`

`2x^2 = -1 \ -> \ text{no real solutions}`

`text{Substitute} \ \ x= -y \ \ text{into} \ (2)`

`-2x^2`	`= -1`
`x`	`= ± 1/sqrt2`

`:. \ z = 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = – 1/sqrt2+ 1/sqrt2 i`

`text{Solution 2}`

`z = re^{i theta} \ , \ z^2 = -i`

`r^2 e^{i2 theta} = e^{i {3pi}/{2}} \ \ text{or} \ \ e^{-i pi/2}`

`=> \ r = 1 \ , \ theta = {3pi}/{4} \ \ text{or} \ \ – pi/4`

`z= text{cos} {3pi}/{4} + i \ text{sin} {3pi}/{4} = – 1/sqrt2 + 1/sqrt2 i`

`z= text{cos} (- pi/4) + i \ text{sin} (- pi/4) = 1/sqrt2 – 1/sqrt2 i`

ii. `z^2 + 2z + 1 + i = 0`

`z`	`= {2 ± sqrt{4 – 4 * 1 * (1 + i)}}/{2}`
	`= {-2 ± sqrt{4 – 4 – 4 \ i}}/{2}`
	`= -1 ± sqrt(-i)`

`:. \ z = -1 + 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = -1 – 1/sqrt2 + 1/sqrt2 i`

Complex Numbers, EXT2 N2 2020 HSC 11e

Solve `z^2 + 3 z + (3-i) = 0`, giving your answer(s) in the form `a + bi`, where `a` and `b` are real. (4 marks)

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`z= -1 + i \ \ text{or} \ \ -2-i`

Show Worked Solution

`z^2 + 3z + (3-i) = 0`

`z`	`= frac{-3 ± sqrt(9-4 · 1 (3-i))}{2}`
	`= frac{-3 ± sqrt(4i-3)}{2} `

`text{Consider} \ \ Delta = sqrt(4i-3) :`

`x + i y`	`= sqrt(4i-3)`
`(x + iy)^2`	`= 4i-3`
`x^2-y^2 + 2xyi`	`= 4i-3`

`text{Equating real and imaginary parts:}`

`2 xy`	`= 4`
`xy`	`= 2\ …\ (1)`
`x^2-y^2`	`= -3\ …\ (2)`

`=> x = 1 \ , \ y =2`

`=> \ x + iy = 1 + 2 i`

`therefore z`	`= frac{-3 ± (1 + 2i)}{2}`
`z `	`= -1 + i \ \ text{or} \ \ -2-i`

Complex Numbers, EXT2 N2 2020 HSC 2 MC

Given that `z = 3 + i` is a root of `z^2 + pz + q = 0`, where `p` and `q` are real, what are the values of `p` and `q`?

`p = -6 \ , \ q = sqrt(10)`
`p = -6 \ , \ q = 10`
`p = 6 \ , \ q = sqrt(10)`
`p = 6 \ , \ q = 10`

Show Answers Only

`B`

Show Worked Solution

`text{S}text{ince} \ \ z_1 = 3 + i \ \ text{is a root}`

`=> \ z_2 = 3 – i \ \ text{is a root}`

`z_1 + z_2`	`= frac{-b}{a}`
`(3 + i) + (3 – i)`	`= -p`
`therefore \ p`	`= -6`

`z_1 z_2`	`= frac{c}{a}`
`(3 + i) (3 – i)`	`= q`
`3^2 – i^2`	`= q`
`therefore \ q`	`= 10`

`=> \ B`

Complex Numbers, EXT2 N1 SM-Bank 4

Express `z` in the form `a + bi`, given `z = sqrt(−3 - 4i)` (2 marks)

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Hence, using the quadratic formula, solve

`z^2 - 7z + 13 + i = 0` (1 mark)

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`z_1 = −1 + 2i`
`z = 3 + i\ \ text(or)\ \ 4 – i`

Show Worked Solution

i.	`z`	`= sqrt(−3 – 4i)`
	`z^2`	`= −3 – 4i`

`−3 – 4i`	`= (x + iy)^2`
	`= x^2 – y^2 + 2xyi`

`x^2 – y^2`	`= −3\ …\ \ (1)`
`2xy`	`= −4`
`xy`	`= −2\ …\ \ (2)`

`x=-1,\ \ y=2`

`x=1,\ \ y=-2`

`:. z_1`	`= −1 + 2i`
`z_2`	`= 1 – 2i`

ii. `z^2 – 7z + 13 + i = 0`

`text(Using general formula:)`

`z`	`= (−b ± sqrt(b^2 – 4ac))/(2a)`
	`= (7 ± sqrt(49 – 4 · 1(13 + i)))/2`
	`= (7 ± sqrt(−3 – 4i))/2`

`text(Using)\ \ z_1 = −1 + 2i,`

COMMENT: Since `z_1 = – z_2`, the general formula only produces 2 distinct solutions.

`z = (7 + (−1 + 2i))/2 = 3+i`

`text(Using)\ \ z_2 = 1 – 2i,`

`z = (7 + (1 – 2i))/2 = 4 – i`

`:. z = 3 + i\ \ text(or)\ \ z=4 – i`

Complex Numbers, EXT2 N2 2017 HSC 12b

Solve the quadratic equation `z^2 + (2 + 3i)z + (1 + 3i) = 0`, giving your answers in the form `a + bi`, where `a` and `b` are real numbers. (3 marks)

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`-1 or -1 – 3i`

Show Worked Solution

`z^2 + (2 + 3i)z + (1 + 3i) = 0`

`z`	`= (-(2 + 3i) +- sqrt((2 + 3i)^2 – 4 · 1(1 + 3i)))/2`
	`= (-(2 + 3i) +- sqrt (4 + 12i + 9i^2 – 4 – 12i))/2`
	`= (-(2 + 3i) +- sqrt(-9))/2`
	`= ((-2 – 3i) +- 3i)/2`

`:. z`	`= (-2)/2 or (-2 – 6i)/2`
	`= -1 or -1 – 3i`

Complex Numbers, EXT2 N2 2016 HSC 10 MC

Suppose that `x + 1/x = -1.`

What is the value of `x^2016 + 1/x^2016?`

`1`
`2`
`(2 pi)/3`
`(4 pi)/3`

Show Answers Only

`=> B`

Show Worked Solution

`x + 1/x = −1`

♦ Mean mark 43%.

`x^2 + x + 1 = 0`

`:. x`	`= (−1 ± sqrt(1 – 4))/2`
	`= −1/2 ± sqrt3/2 i`

`|\ x\ | = sqrt((−1/2)^2 + (sqrt3/2)^2) = 1`

`:. x`	`= cos\ (2pi)/3 + isin\ (2pi)/3`
	`= text(cis)(2pi)/3`

`text(or)`

`x`	`= cos\ (2pi)/3 – isin\ (2pi)/3`
	`= text(cis)(−(2pi)/3)`

`x^2016`	`= text(cis)(±1344pi)quad(text(De Moivre))`
	`= text(cis)0`
	`= 1`

`:. x^2016 + 1/(x^2016) = 2`

`=> B`

Complex Numbers, EXT2 N2 2007 HSC 2d

The points `P,Q` and `R` on the Argand diagram represent the complex numbers `z_1, z_2` and `a` respectively.

The triangles `OPR` and `OQR` are equilateral with unit sides, so `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let `omega = cos pi/3 + i sin pi/3.`

Explain why `z_2 = omega a.` (1 mark)

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Show that `z_1 z_2 = a^2.` (1 mark)

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Show that `z_1` and `z_2` are the roots of `z^2 - az + a^2 = 0.` (2 marks)

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`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i. `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

ii. `text(Solution 1)`

`text(Similarly,)\ \ a`	`=z_1 omega`
`z_1`	`=a/omega`
`:z_1z_2`	`=a/omega xx omega a`
	`=a^2`

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1`	`= bar omega a.`
`:. z_1 z_2`	`= bar omega a xx omega a`
	`=a^2(cos pi/3 – i sin pi/3) xx (cos pi/3 + i sin pi/3)`
	`=a^2(cos^2 pi/3 + sin^2 pi/3)`
	`= a^2`

iii. `z^2-az + a^2 = 0`

`text(Let the roots be)\ \ alpha and beta.`

`alpha + beta`	`=-b/a=a`
`alpha beta`	`=c/a=a^2`

`z_1 z_2`	`= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2`	`=bar omega a + omega a`
	`=(cos pi/3 + i sin pi/3 + cos pi/3-i sin pi/3) a`
	`=2 cos pi/3 xx a`
	`=2 xx 1/2 xx a`
	`=a`

`:.\ z_1 and z_2\ \ text(are the roots of)\ \ z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N2 2009 HSC 2f

Find the square roots of `3 +4i.` (3 marks)

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Hence, or otherwise, solve the equation

`z^2 + iz - 1 - i = 0.` (2 marks)

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`+-(2 + i)`
`1 or -1-i`

Show Worked Solution

i.	`z`	`=sqrt(3+4i)`
	`z^2`	`=3+4i`
	`(x + iy)^2`	`= 3 + 4i`
	`x^2 – y^2+ 2xyi`	`= 3 + 4i`
`=>x^2 – y^2 = 3,\ \ \ xy = 2`

`text(By inspection,)`

`text(If)\ \ x=2,\ \ \ y=1`

`text(If)\ \ x=-2,\ \ \ y=-1`

`:.z= 2 + i,\ \ text(or)\ \ -(2+i)`

ii. `z^2 + iz – 1 – i = 0`

`:.z =`	`(-i +- sqrt (-1 + 4 (1 + i)))/2`
`=`	`\ \ (-i +- sqrt(3 + 4i))/2`
`=`	`\ \ (-i +- (2+i))/2`
`=`	`\ \ 1\ \ \ text(or)\ \ \ -1-i`

Complex Numbers, EXT2 N1 2013 HSC 11c

Factorise `z^2 + 4iz + 5.` (2 marks)

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`(z – i) (z + 5i)`

Show Worked Solution

`text(Solution 1)`

`alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2`

`:. z^2 + 4iz + 5 = (z – i) (z + 5i)`

`(alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2)`

MARKER’S COMMENT: Best practice includes stating the general quadratic formula before substituting in values.

`text(Solution 2)`

`z`	`=(-b+- sqrt(b^2 – 4ac))/(2a)`
	`=(-4i +- sqrt((4i)^2-4 xx 5))/2`
	`=(-4i+-sqrt(-16-20))/2`
	`=(-4i +- 6i)/2`
	`=i\ \ text(or)\ \ -5i`

`:.z^2 + 4iz + 5 = (z – i) (z + 5i)`