Geometric Representations

Complex Numbers, EXT2 EQ-Bank 15

Prove that for any complex numbers \(z_1\) and \(z_2\),
\(\abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}\) (2 marks)
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Hence, or otherwise, show that if \(\abs{z-1}+\abs{z+1} \leqslant 4\) for \(z\in C,\)
\(\abs{z} \leqslant 2\) (2 marks)

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a. \(\text{Proof (See Worked Solutions)}\)

b. \(\text{See Worked Solutions}\)

Show Worked Solution

a. \(\text {Prove}\ \ \abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}:\)

\(\abs{z_1+z_2}^2\)	\(=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)\)
	\(=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2\)
	\(=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)\)

\(\text{Since}\ \ \operatorname{Re}(w) \leqslant\abs{w}\ \ \text{for} \ \ w\in C,\)

\(\abs{z_1+z_2}^2\)	\(\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}\)
	\(\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2\)
	\(\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2\)

\(\therefore\abs{z_1+z_2} \leqslant\abs{z_1}+\abs{z_2}\)

b. \(|z-1|+|z+1| \leqslant 4 \ \text{(given)}\ …\ (1)\)

\(\text {Using triangle inequality:}\)

\(|(z-1)+(z+1)| \leqslant|z-1|+|z+1|\)

\(\text{Since}\ \ (z-1)(z+1)=2 z:\)

\(\abs{2z}\)	\(\leqslant\abs{z-1}+\abs{z+1}\)
\(\abs{2z}\)	\(\leqslant 4\ \ \text{(using (1) above)}\)
\(2\abs{z}\)	\(\leqslant 4\)
\(\abs{z}\)	\(\leqslant 2\)

Complex Numbers, EXT2 N2 2023 14a*

Let \(z\) be the complex number \(z=\text{cis}\dfrac{\pi}{6} \) and \(w\) be the complex number \(w=\text{cis}\dfrac{3\pi}{4} \).

By first writing \(z\) and \(w\) in Cartesian form, or otherwise, show that
\(|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}\). (3 marks)

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The complex numbers \(z, w\) and \(z+w\) are represented in the complex plane by the vectors \(\overrightarrow{O A},\overrightarrow{O B}\) and \(\overrightarrow{O C}\) respectively, where \(O\) is the origin.
Show that \(\angle A O C=\dfrac{7 \pi}{24}\). (2 marks)

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Deduce that \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\). (1 mark)

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i. \(\text{See Worked Solutions}\)

ii. \(\text{See Worked Solutions}\)

iii. \(\text{See Worked Solutions}\)

Show Worked Solution

i. \(z= \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \)

\(w= \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \)

\(\|z+w\|^2\)	\(=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}\)
	\(=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}\)
	\(=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}\)
	\(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)
	\(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)
	\(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)
	\(=\dfrac{4-\sqrt6+\sqrt2}{2} \)

ii.

\(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \)

\( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \)

\(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \)

\(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \)

\(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \)

iii. \(\text{In}\ \triangle AOC: \)

\( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \)

\(\text{Using the cos rule in}\ \triangle AOC: \)

\(\cos\,\dfrac{7\pi}{24}\)	\(=\dfrac{\|z\|^2+\|z+w\|^2-\|w\|^2}{2\|z\|\|z+w\|}\)
	\(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)
	\(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)
	\(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)
	\(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)

♦♦ Mean mark (iii) 26%.

Complex Numbers, EXT2 N1 2025 HSC 7 MC

The complex number \(z\) lies on the unit circle.

What is the range of \(\operatorname{Arg}(z-2 i)\) ?

\(\dfrac{\pi}{6} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{5 \pi}{6}\)
\(\dfrac{\pi}{3} \leq \operatorname{Arg}(z-2 i) \leq \dfrac{2 \pi}{3}\)
\(-\dfrac{5 \pi}{6} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{6}\)
\(-\dfrac{2 \pi}{3} \leq \operatorname{Arg}(z-2 i) \leq-\dfrac{\pi}{3}\)

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\(D\)

Show Worked Solution

\(\text{The limits of Arg}(z-2i)\ \text{where it intersects the unit circle are:}\)

♦ Mean mark 52%.

\(\sin \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text {Range Arg}(z-2 i):\)

\(-\dfrac{\pi}{2}-\dfrac{\pi}{6} \leqslant \operatorname{Arg}(z-2 i) \leqslant-\dfrac{\pi}{2}+\dfrac{\pi}{6}\)

\(-\dfrac{2 \pi}{3} \leqslant \operatorname{Arg}(2-2 i) \leqslant-\dfrac{\pi}{3}\)

\(\Rightarrow D\)

Complex Numbers, EXT2 N2 2025 HSC 6 MC

The complex numbers \(z\) and \(w\) lie on the unit circle. The modulus of \(z+w\) is \(\dfrac{3}{2}\).

What is the modulus of \(z-w\) ?

\(\dfrac{1}{8}\)
\(\dfrac{\sqrt{7}}{2}\)
\(\dfrac{3}{2}\)
\(\dfrac{7}{4}\)

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\(B\)

Show Worked Solution

\(|z|=|w|=1, \quad \abs{z+w}=\dfrac{3}{2} \ \text{(given)}\)

\(\text{Find}\ \ \abs{z-w}:\)

\(\abs{z+\omega}^2=(z+\omega)(\bar{z}+\bar{\omega})=z \bar{z}+z \bar{\omega}+\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z-\omega}^2=(z-\omega)(\bar{z}-\bar{\omega})=\bar{z} z-z \bar{\omega}-\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z+\omega}^2+\|z-\omega\|^2\)	\(=2 z \bar{z}+2 \omega \bar{\omega}=2\abs{z}^2+2\abs{\omega}^2=4\)
\(\dfrac{9}{4}+\abs{z-\omega}^2\)	\(=4\)
\(\abs{z-w}^2\)	\(=\dfrac{7}{4}\)
\(\abs{z-\omega}\)	\(=\dfrac{\sqrt{7}}{2}\)

\(\Rightarrow B\)

Mean mark 56%.

Complex Numbers, EXT2 N1 2024 VCAA 5 MC

If the point \(z=1+\sqrt{3} i\) is represented on an Argand diagram, the point representing \(-\bar{z}\) can be located by

reflecting the point representing \(z\) in the real axis.
rotating the point representing \(z\) anticlockwise about the origin by 90\(^{\circ}\).
reflecting the point representing \(z\) in the imaginary axis.
rotating the point representing \(z\) clockwise about the origin by 90\(^{\circ}\).

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\(C\)

Show Worked Solution

\(z=1+\sqrt{3}i\ \ \Rightarrow\ \ \bar{z} = 1-\sqrt{3}i\)

\(-\bar{z} = -1+\sqrt{3}i\)

\(\therefore \text{It is a reflection of}\ z\ \text{in the imaginary axis.}\)

\(\Rightarrow C\)

Complex Numbers, EXT2 N2 2024 HSC 14c

For the complex numbers \(z\) and \(w\), it is known that \(\arg \left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}\).

Find \(\left|\dfrac{z-w}{z+w}\right|\). (2 marks)

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\(\abs{\dfrac{z-w}{z+w}}=1\)

Show Worked Solution

\(\arg \left(\dfrac{z}{w}\right)=\arg \, z-\arg \, w=-\dfrac{\pi}{2}\)

\(\text{Graphically, \(\arg \, z\) is a \(90^{\circ}\) anticlockwise rotation from \(\arg \, w\).}\)

♦ Mean mark 49%.

\(\abs{z-w}=\abs{z+w} \ \text{(diagonals of rectangle)}\)

\(\therefore \abs{\dfrac{z-w}{z+w}}=1\)

Complex Numbers, EXT2 N1 2022 HSC 15d

The complex number `z` satisfies `|z-(4)/(z)|=2`.

Using the triangle inequality, or otherwise, show that `|z| <= sqrt5+1`. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

♦♦♦ Mean mark 27%.

`text{Triangle Inequality:}\ \ absx+absy>=abs(x+y)`

`absz`	`<=abs(z-z/4)+abs(4/z)`
`absz`	`<=2+4/absz\ \ \ (text{using}\ \|z-(4)/(z)\|=2)`
`absz^2`	`<=2absz+4`
`absz^2-2absz-4`	`<=0`
`absz`	`<=(2+sqrt(2^2+4xx4))/2\ \ \ (absz>=0)`
`absz`	`<=(2+sqrt20)/2`
`absz`	`<=1+sqrt5\ \ text{… as required}`

Complex Numbers, EXT2 N1 2021 SPEC2 4 MC

For the complex number `z `, if `text(Im)(z) > 0`, then `text(Arg)((zbarz)/(z - barz))` is

`-pi/2`
`0`
`pi/4`
`pi`

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`A`

Show Worked Solution

`text(Let)\ \ z=x+iy \ => \ barz=x-iy`

`text(Arg)((zbarz)/(z – barz))`	`= text(Arg)(zbarz) – text(Arg)(z – barz)`
	`= text(Arg)(x^2 + y^2) – text(Arg)(2yi)`
	`= 0 – text(Arg)(2yi),\ \ text(where)\ y > 0`
	`= -pi/2`

`=>\ A`

Complex Numbers, EXT2 N1 2004 HSC 2b

Let `alpha = 1 + i sqrt3` and `beta = 1 + i`.

Find `frac{alpha}{beta}`, in the form `x + i y`. (1 mark)

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Express `alpha` in modulus-argument form. (3 marks)

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Given that `beta` has the modulus-argument form

`beta = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`.

find the modulus-argument form of `frac{alpha}{beta}`. (1 mark)

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Hence find the exact value of `sin frac{pi}{12}` (1 mark)

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a. `frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

b. `2 \ text{cis} (frac{pi}{3})`

c. `sqrt2 \ text{cis} (frac{pi}{12})`

d. `frac{sqrt6-sqrt2}{4}`

Show Worked Solution

a.	`frac{alpha}{beta}`	`= frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}`
		`= frac{(1 + i sqrt3)(1 – i)}{1^2 – i^2}`
		`= frac{1 – i + i sqrt3 – i^2 sqrt3}{2}`
		`= frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

b.	`alpha`	`= 1 + i sqrt3`
	`\| alpha \|`	`= sqrt(1^2 + (sqrt3)^2) = 2`

`text{arg} \ (alpha) = tan^-1 (frac{sqrt3}{1}) = frac{pi}{3}`

`therefore \ alpha = 2 text{cis} (frac{pi}{3})`

c.	`beta`	`= sqrt2 text{cis} (frac{pi}{4})`
	`frac{alpha}{beta}`	`= frac{2}{sqrt2} \ text{cis} (frac{pi}{3} – frac{pi}{4})`
		`= sqrt2 text{cis} (frac{pi}{12})`

d. `text{Equating imaginary parts of i and ii:}`

`sqrt2 \ sin \ (frac{pi}{12})`	`= frac{sqrt3 – 1}{2}`
`sin (frac{pi}{12})`	`= frac{sqrt3 – 1}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
	`= frac{sqrt6 – sqrt2}{4}`

Complex Numbers, EXT2 N1 2008 HSC 2b

Write `frac{1 + i sqrt3}{1 + i}` in the form `x + iy`, where `x` and `y` are real. (2 marks)

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By expressing both `1 + i sqrt3` and `1 + i` in modulus-argument form, write `frac{1 + i sqrt3}{1 + i}` in modulus-argument form. (3 marks)

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Hence find `cos frac{pi}{12}` in surd form. (1 mark)

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a. `frac{1 + sqrt3}{2}-i ( frac{1-sqrt3}{2} )`

b. `sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`

c. `frac{sqrt2 + sqrt6}{4}`

Show Worked Solution

a.	`frac{1 + i sqrt3}{1 + i} xx frac{1-i}{1-i}`	`= frac{(1 + i sqrt3)(1-i)}{1-i^2}`
		`= frac{1-i + i sqrt3-sqrt3 i^2}{2}`
		`= frac{1 + sqrt3}{2}-i ( frac{1-sqrt3}{2} )`

b. `z_1 = 1 + i sqrt3`

`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`

`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`

`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`

`z_2 = 1 + i`

`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`

`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`

`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`

`frac{1 + i sqrt3}{1 + i}`	`= frac{z_1}{z_2}`
	`= frac{2}{sqrt2} ( cos ( frac{pi}{3}-frac{pi}{4} ) + i sin ( frac{pi}{3}-frac{pi}{4} ) )`
	`= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )`

c. `text{Equating real parts of i and ii:}`

`sqrt2 cos (frac{pi}{12})`	`= frac{1 + sqrt3}{2}`
`cos(frac{pi}{12})`	`= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}= frac{sqrt2 + sqrt6}{4}`

Complex Numbers, EXT2 N1 2020 HSC 4 MC

The diagram shows the complex number `z` on the Argand diagram.

Which of the following diagrams best shows the position of `frac{z^2}{|z|}`?

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`A`

Show Worked Solution

`text{Let} \ \ z = r\ text(cis)\ theta`

`z^2`	`= r^2 text(cis)\ (2 theta)`
`\|z\|`	`= r`
`therefore frac{z^2}{\|z\|}`	`= frac{r^2 \ text(cis)\ (2 theta)}{r}`
	`= r\ text(cis)\ (2 theta)`

`text{On Argand diagram, it lies on the dotted line`

`text{(modulus the same) with an argument that is}`

`text{doubled.}`

`=> \ A`

Complex Numbers, EXT2 N1 2018 HSC 13b

Let `z = 1 - cos2theta + isin2theta`, where `0 < theta <= pi`.

Show that `|\ z\ | = 2sintheta`. (2 marks)

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Show that `text(arg)(z) = pi/2 - theta`. (2 marks)

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`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `z = 1 – cos2theta + isin2theta`

`\|\ z\ \|`	`= sqrt((1 – cos2theta)^2 + sin^2 2theta)`
	`= sqrt(1 – 2cos2theta + cos^2 2theta + sin^2 2theta)`
	`= sqrt(2 – 2cos2theta)`
	`= sqrt(2(1 – cos2theta))`
	`= sqrt(2(2sin^2theta))`
	`= 2sintheta\ \ \ text(… as required)`

ii. `z= 1 – cos2theta + isin2theta`

`text(arg)(z)`	`= tan^(−1)((sin2theta)/(1 – cos2theta))`
	`= tan^(−1)((2sinthetacostheta)/(2sin^2theta))`
	`= tan^(−1)(cottheta)`
	`= tan^(−1)(tan(pi/2 – theta))`
	`= pi/2 – theta`

`(text(S)text(ince)\ \ 0 < theta < pi => \ −pi/2 <= pi/2 – theta < pi/2)`

Complex Numbers, EXT2 N1 2017 HSC 11a

Let `z = 1 - sqrt 3 i` and `w = 1 + i`.

Find the exact value of the argument of `z`. (1 mark)
Find the exact value of the argument of `z/w`. (2 marks)

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`-pi/3`
`-(7 pi)/12`

Show Worked Solution

i. `z = 1 – i sqrt 3`

`text(arg)\ z = – pi/3`

ii. `w = 1 + i`

`text(arg)\ w = pi/4`

`text(arg)\ (z/w)`	`= text(arg)\ (z) – text(arg)\ (w)`
	`= – pi/3 – pi/4`
	`= -(7 pi)/12`

Complex Numbers, EXT2 N1 2016 HSC 4 MC

The Argand diagram shows the complex numbers `z` and `w`, where `z` lies in the first quadrant and `w` lies in the second quadrant.

Which complex number could lie in the 3rd quadrant?

`-w`
`2 iz`
`bar z`
`w - z`

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`=> D`

Show Worked Solution

`text(Using the parallelogram method:)`

`text(From the graph above,)`

`w-z\ \ text(could lie in 3rd quadrant.)`

`=> D`

Complex Numbers, EXT2 N1 2015 HSC 12a

The complex number `z` is such that `|\ z\ |=2` and `text(arg)(z) = pi/4.`

Plot each of the following complex numbers on the same half-page Argand diagram.

`z` (1 mark)

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`u = z^2` (1 mark)

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`v = z^2-bar z` (1 mark)

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a. `text(See Worked Solutions)`

b. `text(See Worked Solutions)`

c. `text(See Worked Solutions)`

Show Worked Solution

a. `z =2 text(cis) pi/4=sqrt 2 (1 + i)`

b. `u=z^2=4 text(cis)\ pi/2=4i`

COMMENT: Mean mark (c) 55%.

c. `v `	`=z^2-bar z`
	`=4i-sqrt 2 (1-i)`
	`=- sqrt 2 + (4 + sqrt 2) i`

Complex Numbers, EXT2 N1 2015 HSC 11b

Consider the complex numbers `z = -sqrt 3 + i` and `w = 3 (cos\ pi/7 + i sin\ pi/7).`

Evaluate `|\ z\ |.` (1 mark)

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Evaluate `text(arg)(z).` (1 mark)

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Find the argument of `z/w.` (1 mark)

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a. `2`

b. `(5 pi)/6`

c. `(29 pi)/42`

Show Worked Solution

a. `\|\ z\ \|`	`= sqrt ((-sqrt3)^2 + 1^2)`
	`= 2`

b. `text(arg)\ (z) =`	`tan^-1 (1- sqrt 3)`
`=`	`pi – pi/6`
`=`	`(5 pi)/6`

c. `text(arg) (z/w) =`	`text(arg)\ z – text(arg)\ w`
`=`	`(5 pi)/6 – pi/7`
`=`	`(29 pi)/42`

Complex Numbers, EXT2 N1 2014 HSC 8 MC

The Argand diagram shows the complex numbers `w`, `z` and `u`, where `w` lies in the first quadrant, `z` lies in the second quadrant and `u` lies on the negative real axis.

Which statement could be true?

`u = zw` and `u = z + w`
`u = zw` and `u = z − w`
`z = uw` and `u = z + w`
`z = uw` and `u = z − w`

Show Answers Only

`B`

Show Worked Solution

`text(Using the parallelogram method, it could be true that)`

`u = z-w`

`=>\ text{Eliminate (A) and (C)}`

`text{arg}(u)`	`=\ text{arg}(w) + text{arg}(z)`
`u`	` = zw`

`=> B`

Complex Numbers, EXT2 N1 2014 HSC 4 MC

Given `z = 2(cos\ pi/3 + i sin\ pi/3)`, which expression is equal to `(bar {:z:})^(−1)`?

`1/2(cos\ pi/3 − i sin\ pi/3)`
`2(cos\ pi/3 − i sin\ pi/3)`
`1/2(cos\ pi/3 + i sin\ pi/3)`
`2(cos\ pi/3 + i sin\ pi/3)`

Show Answers Only

`C`

Show Worked Solution

`z`	`= 2text(cis)(pi/3)`
`barz`	`= 2text(cis)(−pi/3)`
`(barz)^(−1)`	`= (2text(cis)(−pi/3))^(−1)`
	`= 2^(−1)text(cis)(pi/3)`
	`= 1/2text(cis)(pi/3)`
	`= 1/2(cos\ pi/3 + i sin\ pi/3)`

`=> C`

Complex Numbers, EXT2 N1 2013 HSC 3 MC

The Argand diagram below shows the complex number `z.`

Which diagram best represents `z^2?`

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`D`

Show Worked Solution

`text(Consider)\ \ z\ \ text(in polar form:)`

`\|\ z\ \|`	`< 1`
`:.\|\ z^2\ \|`	`< \|\ z\ \|`

`text(arg) (z^2) = 2text(arg) (z)`

`=> D`

Complex Numbers, EXT2 N1 2012 HSC 11d

Write `z = sqrt3-i` in modulus-argument form. (2 marks)

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Hence express `z^9` in the form `x + iy`, where `x` and `y` are real. (1 mark)

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a. `2\ text(cis)(-pi/6)`

b. `i512`

Show Worked Solution

a.	`z`	`=sqrt3-i`
	`\|\ z\ \|`	`=sqrt((sqrt3)^2+1^2)=2`

`:.z = sqrt3 − i`	`= 2(sqrt3/2 − 1/2i)`
	`= 2(cos\ (-pi/6) + i\ sin\ (-pi/6))`
	`= 2\ text(cis)(-pi/6)`

b.	`z^9`	`= 2^9\ (cos\ (-pi/6) + i\ sin\ (-pi/6))^9`
		`= 2^9\ text(cis)(-(9pi)/6)\ \ \ \ text{(by De Moivre)}`
		`=512\ text(cis)(-(3pi)/2)`
		`= 512(0 + i)`
		`=i512`

Complex Numbers, EXT2 N1 2014 HSC 11a

Consider the complex numbers `z = -2-2i` and `w = 3 + i`.

Express `z + w` in modulus–argument form. (2 marks)

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

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a. `sqrt2\ text(cis)(-pi/4)`

b. `−4/5-2/5i`

Show Worked Solution

a. `z + w= −2-2i + 3 + i= 1-i`

	`\|\ z+w\ \|`	`= sqrt2`
	`text(arg)\ (z+w)`	`=- pi/4`
	`:. z+w`	`= sqrt2\ text(cis)(-pi/4)`

b.	`z/w`	`= (−2-2i)/(3 + i)`
		`= ((−2-2i)(3-i))/((3 + i)(3-i))`
		`= (−6-6i + 2i-2)/(9 + 1)`
		`= −8/10 −4/10i`
		`= −4/5-2/5i`

`\|\ z\ \|`	`< 1`
`:.\|\ z^2\ \|`	`< \|\ z\ \|`