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Complex Numbers, EXT2 N1 2025 SPEC2 6*

Let  \(z \in C\).

Given that  \(|z|=1\)  and  \(z \neq 1,\) express  \(\operatorname{Re}\left(\dfrac{1}{1-z}\right)\) in its simplest form.   (3 marks)

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\(\operatorname{Re}\left(\dfrac{1}{1-z}\right) = \dfrac{1}{2}\)

Show Worked Solution

\(z=a+b i\)

\(\abs{z}=a^2+b^2=1\)

\(\dfrac{1}{1-z}=\dfrac{1}{1-(a+bi)}=\dfrac{1}{(1-a)-bi} \times \dfrac{(1-a)+bi}{(1-a)+bi}=\dfrac{(1-a)+bi}{(1-a)^2+b^2}\)

\(\operatorname{Re}\left(\dfrac{1}{1-z}\right)\) \(=\dfrac{-a+1}{a^2-2 a+b^2+1}\)
  \(=\dfrac{-a+1}{1-2 a+1}\)
  \(=\dfrac{-(a-1)}{-2(a-1)}\)
  \(=\dfrac{1}{2}\)

Complex Numbers, EXT2 N1 2025 HSC 3 MC

What are the square roots of  \(3-4 i\) ?

  1. \(1-2 i\)  and  \(-1+2 i\)
  2. \(1+2 i\)  and  \(-1-2 i\)
  3. \(2-i\)  and  \(-2+i\)
  4. \(-2-i\)  and  \(2+i\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Let} \ \ z=\sqrt{3-4 i}\) :

\(z^2=3-4 i\)

\(z^2=a^2-b^2+2 a b\,i\)

\(\text{Equate real/imaginary parts:}\)

\(a^2-b^2=3\ \ldots\ (1)\)

\(2 a b=-4 \ \ \Rightarrow \ \ a b=-2\ \ldots\ (2)\)

\(\text{By inspection:}\)

\(a=2, b=-1 \ \Rightarrow \ z_1=2-i\)

\(a=-2, b=1 \ \Rightarrow \ z_2=-2+i\)

\(\Rightarrow C\)

Complex Numbers, EXT2 N1 2025 HSC 11c

The complex number \(z\) is given by  \(x+i y\).

Find, in Cartesian form:

  1. \(z^2\)   (1 mark)

  2. \(\dfrac{1}{z}\).   (2 marks)

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i.    \(z^2=x^2-y^2+2xy\,i\)

ii.  \(\dfrac{1}{z}=\dfrac{x}{x^2-y^2}-\dfrac{y}{x^2-y^2}\,i\)

Show Worked Solution
i.    \(z^{2}\) \(=(x+iy)^2\)
    \(=x^2-y^2+2xy\,i\)

 

ii.    \(\dfrac{1}{z}\) \(=\dfrac{1}{x+iy}\)
    \(=\dfrac{x-iy}{(x+iy)(x-iy)}\)
    \(=\dfrac{x-iy}{x^2+y^2}\)
    \(=\dfrac{x}{x^2+y^2}-\dfrac{y}{x^2+y^2}\,i\)

Complex Numbers, EXT2 N1 2024 HSC 12c

Consider the equation  \(\abs{z}=z+8+12 i\), where  \(z=a+b i\)  is a complex number and \(a, b\) are real numbers.

  1. Explain why  \(b=-12\).   (1 mark)

  2. Hence, or otherwise, find \(z\).   (2 marks)

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i.     \(\abs{z}=z+8+12i\)

\(z=a+b i\ \ \Rightarrow\ \ \abs{z}=\sqrt{a^2+b^2}\)

\(\text{Equating moduli:}\)

\(\sqrt{a^2+b^2}=a+b i+8+12 i=a+8+(b+12) i\)

\(\text{Since } \sqrt{a^2+b^2} \in \mathbb{R}:\)

\(b+12=0 \ \Rightarrow \ b=-12\)
 

ii.    \(z=5-12 i\)

Show Worked Solution

i.     \(\abs{z}=z+8+12i\)

\(z=a+b i\ \ \Rightarrow\ \ \abs{z}=\sqrt{a^2+b^2}\)

\(\text{Equating moduli:}\)

\(\sqrt{a^2+b^2}=a+b i+8+12 i=a+8+(b+12) i\)

\(\text{Since } \sqrt{a^2+b^2} \in \mathbb{R}:\)

\(b+12=0 \ \Rightarrow \ b=-12\)
  

ii.    \(\abs{a-12 i}\) \(=a+8\)
  \(\sqrt{a^2+144}\) \(=a+8\)
  \(a^2+144\) \(=a^2+16 a+64\)
  \(16a\) \(=80\)
  \(a\) \(=5\)
 
\(\therefore z=5-12 i\)

Complex Numbers, EXT2 N1 2024 HSC 11b

Let  \(z=2+3 i\)  and  \(w=1-5 i\).

  1. Find  \(z+\bar{w}\).   (1 mark)

  2. Find  \(z^2\).   (1 mark)

Show Answers Only

i.    \(z+\bar{w}=3+8 i\)

ii.   \(z^2=-5+12 i\)

Show Worked Solution

i.     \(z=2+3 i\)

\(w=1-5 i \ \Rightarrow \ \bar{w}=1+5 i\)

\(z+\bar{w}=2+3 i+1+5 i=3+8 i\)
 

ii.    \(z^2\) \(=(2+3 i)^{2}\)
    \(=4+12 i+9 i^2\)
    \(=-5+12 i\)

Complex Numbers, EXT2 N1 2023 HSC 1 MC

Which of the following is equal to \((a+i b)^3\)?

  1. \( (a^3-3 a b^2)+i (3 a^2 b+b^3) \)
  2. \( (a^3+3 a b^2)+i (3 a^2 b+b^3) \)
  3. \( (a^3-3 a b^2)+i (3 a^2 b-b^3) \)
  4. \( (a^3+3 a b^2)+i(3 a^2 b-b^3)\)
Show Answers Only

\(C\)

Show Worked Solution
\((a+i b)^3\) \(=a^3+3a^2ib+3a(ib)^2+(ib)^3\)  
  \(=a^3+3a^2ib-3ab^2-ib^3\)  
  \(=(a^3-3ab^2)+i(3a^2b-b^3) \)  

 
\(\Rightarrow C\)

Complex Numbers, EXT2 N1 2021 HSC 11e

The complex numbers  `z = 5 + i`  and  `w = 2 − 4 i`  are given.

Find  `bar z/{w}`, giving your answer in Cartesian form.  (2 marks)

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`7/10 + 9/10 i`

Show Worked Solution

`z = 5 + i \ => \ bar z= 5 – i`

`w = 2- 4 i`

`bar z/w` `= {5 – i}/{2 – 4 i} xx {2 + 4 i}/{2 + 4 i}`
  `= {(5 – i)(2 +  4i)}/{4 + 16}`
  `= {10 + 20 i – 2i + 4}/{20}`
  `= 7/10 + 9/10 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)

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

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

Show Worked Solution

i.    `text{Method 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{Method 2 (exponential form – ex-syllabus in 2027):}`

`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 N1 2020 SPEC2 8 MC

Given that  `(x + iy)^14 = a + ib`, where  `x, y, a, b ∈ R, \ (y - ix)^14`  for all values of `x` and `y` is equal to

  1. `−a - ib`
  2. `−b + ia`
  3. `−a + ib`
  4. `b + ia`
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`A`

Show Worked Solution
`(y – ix)^14` `= (−i(x + iy))^14`
  `= −i^14(x + iy)^14`
  `= −(x + iy)^14`
  `= −a – ib`

 
`=>A`

Complex Numbers, EXT2 N2 2020 SPEC2 5 MC

Given the complex number  `z = a + bi`, where  `a ∈ R text{\}{0}`  and  `b ∈ R, \ (4zbarz)/((z + barz)^2)` is equivalent to

  1. `1 + ((text(Im)(z))/(text(Re)(z)))^2`
  2. `4[text(Re)(z) xx text(Im)(z)]`
  3. `4([text(Re)(z)]^2 + [text(Im)(z)]^2)`
  4. `(2 xx text(Im)(z))/([text(Re)(z)]^2)`
Show Answers Only

`A`

Show Worked Solution

`z = a + ib, \ barz = a – ib`

`(4zbarz)/((z + barz)^2)` `= (4(a^2 + b^2))/(4a^2)`
  `= 1 + (b/a)^2`
  `= 1 + ((text(Im)(z))/(text(Re)(z)))^2`

`=>A`

Complex Numbers, EXT1 N1 EQ-Bank 11

Let  `z = 1 + 2 i`  and  `w = 3-i`.

Find, in the form  `x + i y`,

  1.  `zw`   (1 mark)

  2.  `overset_((frac{10}{z}))`.   (1 mark)

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a.    `5 + 5 i`

b.   `2+4 i`

Show Worked Solution
a.     `zw` `= (1 + 2 i)(3-i)`
    `= 3-i + 6 i-2 i^2`
    `= 5 + 5 i`

 

b.    `frac{10}{z}` `= frac{10}{1 + 2 i} xx frac{1-2 i}{1-2 i}`
    `= frac{10-20 i}{1^2-(2 i)^2`
    `= frac{10-20 i}{1+4}`
    `= 2-4i`

 
`therefore \ overset_(frac{10}{z})= 2+4 i`

Complex Numbers, EXT2 N1 2005 HSC 2a

Let  `z = 3 + i`  and  `w=1-i`.  Find, in the form  `x+iy`,

  1.  `2z+iw`   (1 mark)

  2.  `bar z w`   (1 mark)

  3.  `6/w`   (1 mark)

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a.    `7+3i`

b.    `2-4i`

c.    `3+3i`

Show Worked Solution

a.    `z = 3 + i \ , \ w = 1 – i`

`2z + i w` `=2 (3 + i) + i(1 – i)`
  `= 6 + 2i + i – i^2`
  `= 7 + 3 i`

 

b.     `overset_z w` `= (3 – i)(1 – i)`
    `= 3 – 3i –  i + i^2`
    `= 2 – 4 i`

 

c.     `frac{6}{w}` `= frac{6}{1 – i} xx frac{1 + i}{1 + i}`
    `= frac{6 + 6i}{1^2 – i^2}`
    `= frac{6 + 6i}{2}`
    `= 3 + 3i`

Complex Numbers, EXT2 N1 2020 HSC 11a

Consider the complex numbers  `w = -1 + 4i`  and  `z = 2 -i`.

  1. Evaluate  `|w|`.   (1 mark)

  2. Evaluate  `w overset_ z`.   (2 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
Show Worked Solution
i.      `w` ` = -1 + 4 i`
  `|w| ` `= sqrt{(-1)^2 + 4^2} = 17`

 

ii.     `z = 2 – i`

`overset_z = 2 + i`

`w overset_z` `= (-1 + 4i)(2 + i)`
  `= -2 – i +8 i + 4i^2`
  `= -6 + 7i`

Complex Numbers, EXT2 N1 EQ-Bank 13

Find the values of `z`, in the form  `z = x + iy`, such that

`z = sqrt(-15 + 8i)`   (2 marks)

Show Answers Only

`z= 1 + 4i\ \ text(or)\ \ z= -1-4i`

Show Worked Solution
`z` `= sqrt(-15 + 8i)`
`z^2` `= -15 + 8i`
`-15 + 8i` `= (x + iy)^2`
  `= x^2-y^2 + 2xyi`

 

`x^2-y^2` `= -15\ …\ (1)`
`2xy` `= 8`
`xy` `= 4\ …\ (2)`

 
`x=1,\ \ y=4`

`x=-1,\ \ y=-4`
 

`:. z_1` `= 1 + 4i`
`z_2` `= -1-4i`

Complex Numbers, EXT2 N1 EQ-Bank 12

Find the values of `z`, in the form  `z = a + ib`, such that

`z = sqrt(7 + 24i)`   (2 marks)

Show Answers Only

`z = 4 + 3i\ \ text(or)\ \ z=-4-3i`

Show Worked Solution
`z` `= sqrt(7 + 24i)`
`z^2` `= 7 + 24i`
`7 + 24i` `= (a + ib)^2`
  `= a^2-b^2 + 2abi`

 

`a^2-b^2` `= 7\ …\ (1)`
`2ab` `= 24`
`ab` `= 12\ …\ (2)`

 
`a=4,\ \ b=3`

`a=-4,\ \ b=-3`
 

`:. z_1` `= 4 + 3i`
`z_2` `= −4-3i`

Complex Numbers, EXT2 N1 2019 HSC 11a

Let  `z = 1 + 3i`  and  `w = 2 - i`.

  1. Find  `z + bar w`.  (1 mark)

  2. Express  `z/w`  in the form  `x + iy`, where  `x` and  `y`  are real numbers.  (2 marks)

Show Answers Only
  1. `3 + 4i`
  2. `-1/5 + 7/5 i`
Show Worked Solution

i.   `z = 1 + 3i`

`w = 2 – i \ => \ bar w = 2 + i`

`z + bar w` `= 1 + 3i + 2 + i`
  `= 3 + 4i`

 

ii.    `z/w` `= (1 + 3i)/(2 – i) xx (2 + i)/(2 + i)`
    `= ((1 + 3i)(2 + i))/(2^2 – i^2)`
    `= (2 + i + 6i + 3i^2)/5`
    `= (-1 + 7i)/5`
    `= -1/5 + 7/5 i`

Complex Numbers, EXT2 N1 2019 HSC 8 MC

Let `z` be a complex number such that  `z^2 = -i bar z`.

Which of the following is a possible value for `z`?

  1. `1/2-sqrt 3/2 i`
  2. `1/2 + sqrt 3/2 i`
  3. `sqrt 3/2-1/2 i`
  4. `sqrt 3/2 + 1/2 i`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

`text(Consider)\ C:`

`z` `=sqrt3/2-1/2 i`  
`barz` `=sqrt3/2 + 1/2 i`  
`-i barz` `=1/2-sqrt3/2 i`  

 

`z^2` `=(sqrt3/2-1/2 i)^2`  
  `=3/4-2 sqrt3/2 * 1/2 i -1/4`  
  `=1/2-sqrt3/2 i`  

 
`=>C`

 

`text(Solution 2)`

`text(Let)\ \ text(arg)(z)` `= theta`
`text(arg)(z^2)` `= 2 theta`

 

`text(arg)(-i bar z)` `=text(arg)(i bar z)-pi`
  `= text(arg) (bar z)-pi + pi/2`
  `= -theta-pi/2`

 

`:. 2 theta` `= -theta-pi/2`
`3 theta` `= -pi/2`
`theta` `= -pi/6`

 
`=>   C`

Complex Numbers, EXT2 N1 2018 HSC 11a

Let  `z = 2 + 3i`  and  `w = 1 - i.`

  1. Find  `zw`.  (1 mark)

  2. Express  `barz  - 2/w`  in the form  `x + iy`, where `x` and `y` are real numbers.  (2 marks)

Show Answers Only
  1. `5 + i`
  2. `1 – 4i`
Show Worked Solution

i.   `z = 2 + 3iqquadw = 1 – i`

♦ Mean mark part (i) 97!%.

`zw` `= (2 + 3i)(1 – i)`
  `= 2 – 2i + 3i – 3i^2`
  `= 2 + i + 3`
  `= 5 + i`

 

ii.    `barz – 2/w` `= 2 – 3i – 2/(1 – i)`
    `= 2 – 3i – (2(1 – i))/((1 – i)(1 + i))`
    `= 2 – 3i – (1 + i)`
    `= 1 – 4i`

Complex Numbers, EXT2 N1 2015 HSC 11a

Express  `(4 + 3i)/(2 - i)`  in the form  `x + iy`, where  `x`  and  `y`  are real.  (2 marks)

Show Answers Only

`1 + 2i`

Show Worked Solution
`(4 + 3i)/(2 – i) ­` `=(4 + 3i)/(2 – i) xx (2 + i)/(2 + i)`
  `=(8 + 4i + 6i – 3)/(4 + 1)`
  `=(5 + 10i)/5`
  `=1 + 2i`

Complex Numbers, EXT2 N1 2007 HSC 2a

Let  `z = 4 + i`  and  `w = bar z`. Find, in the form  `x + iy`,

  1.  `w`   (1 mark)

  2.  `w - z`   (1 mark)

  3.  `z/w`.   (1 mark)

Show Answers Only

a.    `4 – i`

b.    `-2i`

c.    `15/17 + 8/17 i`

Show Worked Solution

a.    `z = 4 + i,\ \ w = bar z = 4 – i`
 

b.    `w – z` `= 4 – i – (4 + i)`
  `= -2i`

 

c.    `z/w` `=(4 + i)/(4 – i) xx (4+i)/(4+i)`
  `=(4 + i)^2/(16+1)`
  `=(16+8i-1)/17`
  `=15/17 + 8/17 i`

Complex Numbers, EXT2 N1 2006 HSC 2a

Let  `z = 3 + i`  and  `w = 2 - 5i`.  Find, in the form  `x + iy`,

  1.  `z^2.`  (1 mark)

  2.  `bar z w.`  (1 mark)

  3.  `w/z.`  (1 mark)

Show Answers Only

a.    `8 + 6i`

b.    `1 – 17i`

c.    `1/10 – 17/10 i`

Show Worked Solution
a.   `z^2` `=(3 + i)^2`
  `= 9 + 6i – 1`
  `= 8 + 6i`

 

b.  `bar{:z:} w` `=(3 – i) (2 – 5i)`
  `= 6 – 15i – 2i – 5`
  `= 1 – 17i`

 

c.  `w/z` `=(2 – 5i)/(3 + i) xx (3 – i)/(3 – i)`
  `= (1 – 17i)/10`
  `= 1/10 – 17/10 i`

Complex Numbers, EXT2 N1 2009 HSC 2b

Write  `(-2 + 3i)/(2 + i)`  in the form  `a + ib`  where  `a`  and  `b`  are real.   (1 mark)

Show Answers Only

`-1/5 + 8/5 i`

Show Worked Solution

`(-2 + 3i)/(2 + i) xx (2-i)/(2-i)`

`= ((-2 + 3i)(2 – i))/(4 + 1)`

`= (-4 + 2i + 6i + 3)/5`

`= (8i – 1)/5`

`= -1/5 + 8/5 i`

Complex Numbers, EXT2 N1 2010 HSC 2a

Let  `z = 5 − i`.

  1. Find  `z^2`  in the form  `x + iy`.   (1 mark)

  2. Find  `z + 2barz`  in the form  `x + iy`.   (1 mark)

  3. Find  `i/z`  in the form  `x + iy`.   (2 marks)

Show Answers Only

a.    `24 − 10i`

b.    `15 + i`

c.    `-1/26 + 5/26 i`

Show Worked Solution
a.     `z` `= 5 − i`
  `:.z^2` `= (5 − i)^2`
    `= 25 − 10i +i\ ^2`
    `= 24 − 10i`

 

b.      `z + 2bar z` `= 5 − i + 2(5 + i)`
    `= 15 + i`

 

c.     `i/z` `= i/(5 − i) xx (5+i)/(5+i)`
    `= (i(5 + i))/(25 + 1)`
    `= (5i − 1)/26`
    `= -1/26 + 5/26 i`

Complex Numbers, EXT2 N1 2011 HSC 2a

Let  `w = 2 - 3i`  and  `z = 3 + 4i.`

  1.  Find  `bar w + z.`   (1 mark)

  2.  Find  `|\ w\ |.`   (1 mark)

  3.  Express  `w/z`  in the form  `a + ib`, where  `a`  and  `b`  are real numbers.   (2 marks)

Show Answers Only

a.    `5 + 7i`

b.    `sqrt13`

c.    `-6/25 -17/25 i`

Show Worked Solution

a.    `w = 2 – 3i,\ \ z = 3 + 4i,\ \ bar w = 2 + 3i`

`bar w + z` `= 2 + 3i + 3 + 4i`
  `= 5 + 7i`

 

b.    `|\ w\ |` `= sqrt (2^2 + 3^2)`
  `= sqrt 13`

 

 c.    `w/z` `= (2 – 3i)/(3 + 4i) xx (3 – 4i)/(3 – 4i)`
  `= (6 – 8i – 9i – 12)/(9 + 16)`
  `= (-6 – 17i)/25`
  `= -6/25 -17/25 i`

Complex Numbers, EXT2 N1 2012 HSC 11a

Express  `(2sqrt5 + i)/(sqrt5 − i)`  in the form  `x + iy`, where `x` and `y` are real.  (2 marks)

Show Answers Only

`3/2 + sqrt5/2 i`

Show Worked Solution
`(2sqrt5 + i)/(sqrt5 − i)` `= ((2sqrt5 + i)(sqrt5 + i))/((sqrt5 − i)(sqrt5 + i))`
  `= (10 + 2sqrt5i + sqrt5i − 1)/(5 + 1)`
  `= (9 + 3sqrt5i)/6`
  `= 3/2 + sqrt5/2 i`