SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Complex Numbers, EXT2 N2 2022 SPEC1 1

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

  1. Express `p(z)` in the form  `p(z) = (z+ai)^2 + b`  where  `a`, ` b  ∈ R`.   (1 mark)
  2. Hence, or otherwise, find the solutions of the equation  `p(z) = 0`.   (2 marks)
Show Answers Only
  1. `p(z) = (z + 3i)^2-16`
  2. `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)

Show Answers Only

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

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.

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

  2. 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)

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

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(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

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

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

Show Answers Only
  1. `z = -1 – i\ \ text(or)\ \ -1 + i`
  2. `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

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

  2. 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)

Show Answers Only
  1. `z = 1/sqrt2 – 1/sqrt2 i \ \ text{or} \ \ z = – 1/sqrt2+ 1/sqrt2 i`
  2. `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)

Show Answers Only

`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`?

  1.  `p = -6 \ , \ q = sqrt(10)`
  2.  `p = -6 \ , \ q = 10`
  3.  `p = 6 \ , \ q = sqrt(10)`
  4.  `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

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

  2. Hence, using the quadratic formula, solve
     
         `z^2 - 7z + 13 + i = 0`  (1 mark)

Show Answers Only
  1. `z_1 = −1 + 2i`
  2. `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)

Show Answers Only

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

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

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

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

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `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

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

  2. Hence, or otherwise, solve the equation

     

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

Show Answers Only
  1. `+-(2 + i)`
  2. `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)

Show Answers Only

`(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)`