smc-2597-20-Cartesian to Mod/Arg

Complex Numbers, SPEC2 2022 VCAA 2

Two complex numbers \(u\) and \(v\) are given by \(u=a+i\) and \(v=b-\sqrt{2}i\), where \(a, b \in R\).

i. Given that \(uv=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6})i\), show that \(a^2+(1-\sqrt{3}) a-\sqrt{3}=0\). (2 marks)

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ii. One set of possible values for \(a\) and \(b\) is \(a=\sqrt{3}\) and \(b=\sqrt{2}\).
Hence, or otherwise, find the other set of possible values. (1 mark)

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Plot and label the points representing \(u=\sqrt{3}+i\) and \(v=\sqrt{2}-\sqrt{2}i\) on the Argand diagram below.

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The ray given by \(\text{Arg}(z)=\theta\) passes through the midpoint of the line interval that joins the points \(u=\sqrt{3}+i\) and \(v=\sqrt{2}-\sqrt{2}i\).
Find, in radians, the value of \(\theta\) and plot this ray on the Argand diagram in part b. (2 marks)

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The line interval that joins the points \(u=\sqrt{3}+i\) and \(v=\sqrt{2}-\sqrt{2}i\) cuts the circle \(|z|=2\) into a major and a minor segment.
Find the area of the minor segment, giving your answer correct to two decimal places. (2 marks)

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Show Answers Only

a.i. \(\text{See worked solutions.}\)

a.ii. \(a=-1, \ b=-\sqrt{6}\)

c. \(\theta=-\dfrac{\pi}{24}\)

d. \(0.69\ \text{u}^2\)

Show Worked Solution

a.i. \(u v=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6}) i\)

\begin{aligned}
(a+i)(b-\sqrt{2} i) & =a b-\sqrt{2} a i+b i+\sqrt{2} \\
& =a b+\sqrt{2}+(b-\sqrt{2}a) i
\end{aligned}

\(\text {Equating co-efficients: }\)

\(ab=\sqrt{6} \ \Rightarrow \ b=\dfrac{\sqrt{6}}{a}\ \cdots\ \text {(1)}\)

\(b-\sqrt{2} a=\sqrt{2}-\sqrt{6}\ \cdots\ \text {(2)}\)

\(\text{Substitute (1) into (2):}\)

\(\dfrac{\sqrt{6}}{a}-\sqrt{2}a=\sqrt{2}-\sqrt{6}\)

\(\sqrt{6}-\sqrt{2} a^2=(\sqrt{2}-\sqrt{6}) a\)

\(\sqrt{2} a^2+(\sqrt{2}-\sqrt{6}) a-\sqrt{6}\)	\(=0\)
\(a^2+(1-\sqrt{3}) a-\sqrt{3}\)	\(=0\)

Mean mark (a) 54%.

a.ii. \(\text{Solve } a^2+(1-\sqrt{3}) a-\sqrt{3}=0 \ \ \text{for}\ a :\)

\(a=\sqrt{3} \ \text{ or } -1\)

\(\therefore \text{ Other solution: } a=-1, \ b=-\sqrt{6}\)

c.	\begin{aligned} \theta & =\frac{\operatorname{Arg}(u)-\operatorname{Arg}(v)}{2} \\ & =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\pi}{4}\right) \\ & =-\frac{\pi}{24} \end{aligned}

♦ Mean mark (c) 39%.

d. \(\text {Sector angle (centre) }=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{5 \pi}{12}\)

\(\text {Area of sector }=\dfrac{\frac{5 \pi}{12}}{2 \pi} \times \pi \times 2^2=\dfrac{5 \pi}{6}\)

\begin{aligned}
\text {Area of segment } & =\frac{5 \pi}{6}-\dfrac{1}{2} \times 2 \times 2 \times \sin \left(\dfrac{5 \pi}{6}\right) \\
& \approx 0.69\ \text{u} ^2
\end{aligned}

♦ Mean mark (d) 40%.

Complex Numbers, SPEC2 2022 VCAA 5 MC

Let `z=x+yi`, where `x, y \in R` and `z \in C`.

If `\text{Arg}(z-i)=\frac{3\pi}{4}`, which one of the following is true?

`y=1-x, x<0`
`y=1-x, x>0`
`y=1+x`
`y=1+x, x>0`
`y=1+x, x<0`

Show Answers Only

`A`

Show Worked Solution

`\text{Arg}(z-i)=\frac{3\pi}{4}`

`\frac{y-1}{x}=tan (\frac{3pi}{4})`

`:.y=1-x, \ x<0`

`=>A`

Complex Numbers, SPEC1 2023 VCAA 2

Consider the complex number \(z=(b-i)^3\), where \(b \in R^{+}\).

Find \(b\) given that \(\arg (z)=-\dfrac{\pi}{2}\). (3 marks)

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Show Answers Only

\(b=\sqrt{3} \)

Show Worked Solution

\(-\dfrac{\pi}{2} = \arg (z) = \arg[(b-i)^3]\)

\(b-i\ \Rightarrow \ \text{complex number that lives on}\ \ y=-i \)

\(3 \times \arg(b-i) = -\dfrac{\pi}{2} \)

\(\arg(b-i) = -\dfrac{\pi}{6} \)

\(\tan^{-1}\Big{(}\dfrac{-1}{b} \Big{)} \)	\(= -\dfrac{\pi}{6} \)
\(-\dfrac{1}{b}\)	\(=-\dfrac{1}{\sqrt{3}}\)
\(\therefore b\)	\(=\sqrt{3}\)

Complex Numbers, SPEC2 2021 VCAA 4 MC

For `z ∈ C`, if `text(Im)(z) > 0`, then `text(Arg)((zbarz)/(z - barz))` is

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

Show Answers Only

`A`

Show Worked Solution

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

♦♦ Mean mark 35%.

`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, SPEC1 2019 VCAA 7

Show that `3 - sqrt3 i = 2sqrt3 text(cis)(−pi/6)`. (1 mark)
Find `(3 - sqrt3 i)^3`, expressing your answer in the form `x + iy`, where `x`, `y ∈ R`. (2 marks)
Find the integer values of `n` for which `(3 - sqrt3 i)^n` is real. (1 mark)
Find the integer values of `n` for which `(3 - sqrt3 i)^n = ai`, where `a` is a real number. (1 mark)

Show Answers Only

`text(See Worked Solutions)`
`0 – i 24sqrt3`
`n = 6k\ \ (k ∈ ZZ)`
`n = 3 + 6k\ \ (k ∈ ZZ)`

Show Worked Solution

a. `|3 – sqrt3 i|= sqrt(3^2 + (−sqrt3)^2)= sqrt12= 2sqrt3`

`text(Arg)(3 – sqrt3 i)`	`= tan^(−1)(−(sqrt3)/3)= −pi/6`

`:. 3 – sqrt3 i = 2sqrt3\ text(cis)(−pi/6)`

b.	`(3 – sqrt3 i)^3`	`= (2sqrt3)^3\ text(cis)(3 xx −pi/6)`
		`= 24sqrt3\ text(cis)(−pi/2)`
		`= 24sqrt3(cos(−pi/2) + isin(−pi/2))`
		`= 0 – i 24sqrt3`

c. `(3 – sqrt3 i)^n = (2sqrt3)^n\ text(cis)((−npi)/6)`

`text(Real when)\ \ sin(−(npi)/6) = −sin((npi)/6) = 0`

`(npi)/6 = 0, pi, 2pi, …, kpi\ \ (k ∈ ZZ)`

`:. n = 6k\ \ (k ∈ ZZ)`

d. `(3 – sqrt3 i)^n = ai\ \ text(when)\ \ cos(−(npi)/6) = cos((npi)/6) = 0`

`(npi)/6 = pi/2, (3pi)/2, …, pi/2 + kpi\ \ (k ∈ ZZ)`

`:. n = 3 + 6k\ \ (k ∈ ZZ)`

Complex Numbers, SPEC2 2012 VCAA 2

Given that `cos(pi/12) = (sqrt (sqrt 3 + 2))/2`, show that `sin(pi/12) = (sqrt (2 - sqrt 3))/2`. (2 marks)
Express `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2 - sqrt3))/2` in polar form. (1 mark)
i. Write down `z_1^4` in polar form. (1 mark)
ii. On the Argand diagram below, shade the region defined by

`{z: text(Arg)(z_1) <= text(Arg)(z) <= text(Arg)(z_1^4)} ∩ {z: 1 <= |\ z\ | <= 2}, z ∈ C`. (2 marks)

Find the area of the shaded region in part c. (2 marks)
i. Find the value(s) of `n` such that `text(Re)(z_1^n) = 0`, where `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2 - sqrt3))/2`. (3 marks)
ii. Find `z_1^n` for the value(s) of `n` found in part i. (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
i. `z_1 = text(cis) pi/12`

ii. `z_1^4 = text(cis) pi/3`
`text(See Worked Solutions.)`
`(3pi)/8`
i. `n = (2k + 1)6`

ii. `z_1^n = ±i\ text(for)\ k ∈ Z`

Show Worked Solution

a.	`cos^2(pi/12) + sin^2(pi/12)`	`= 1`
	`sin^2(pi/12)`	`= 1-(sqrt 3 + 2)/4`
		`= (2 – sqrt 3)/4`

`:. sin (pi/12) = sqrt(2 – sqrt 3)/2,\ \ \ (sin (pi/12) > 0)`

b.i.	`z_1`	`= cos(pi/12) + i sin (pi/12)`
		`= text(cis)(pi/12)`

b.ii.	`z_1^4`	`= 1^4 text(cis) ((4 pi)/12)`
		`= text(cis)(pi/3)`

d. `text(Area of large sector)`

Mean mark 51%.

`=theta/(2pi) xx pi r^2`

`=(pi/3-pi/12)/(2pi) xx pi xx 2^2`

`=pi/2`

`text(Area of small sector)`

`=1/2 xx pi/4 xx 1`

`=pi/8`

`:.\ text(Area shaded)`

`= pi/2 – pi/8`

`= (3 pi)/8`

♦ Mean mark (e)(i) 34%.

e.i.	`z_1^n`	`= 1^n text(cis) ((n pi)/12)`
	`text(Re)(z_1^n)`	`= cos((n pi)/12) = 0`
	`(n pi)/12`	`=cos^(-1)0 +2pik,\ \ \ k in ZZ`
	`(n pi)/12`	`= pi/2 + k pi`
	`n`	`= 6 + 12k,\ \ \ k in ZZ`

e.ii. `text(When)\ \ n=(6 + 12k):`

♦ Mean mark (e)(ii) 36%.

	`z_1^n`	`= 1^(6 + 12k) text(cis) (((6 + 12k)pi)/12), quad k in ZZ`
		`= i xx sin (((6 + 12k)pi)/12)`
		`= i xx sin (pi/2 + k pi)`

`text(If)\ \ k=0 or text(even,)\ \ z_1^n=i`

`text(If)\ \ k\ text(is odd,)\ \ z_1^n=-i`

`:. z_1^n = +-i,\ \ \ k in ZZ`

Complex Numbers, SPEC2 2013 VCAA 8 MC

The principal arguments of the solutions to the equation `z^2 = 1 + i` are

`pi/8` and `(9pi)/8`
`−pi/8` and `(7pi)/8`
`−(7pi)/8` and `pi/8`
`(7pi)/8` and `(15pi)/8`
`−(3pi)/4` and `pi/4`

Show Answers Only

`C`

Show Worked Solution

`z^2`	`= 1 + i`
	`= sqrt2 text(cis)(pi/4 + 2kpi)`
`z`	`= 2^(1/4) text(cis) (pi/8 + kpi)`

`:.\ text(Principal arguments are)\ −(7pi)/8\ text(and)\ pi/8.`

`=> C`

Complex Numbers, SPEC1 2016 VCAA 6

Write `(1 - sqrt 3 i)^4/(1 + sqrt 3 i)` in the form `a + bi`, where `a` and `b` are real constants. (3 marks)

Show Answers Only

`4 + 4 sqrt 3 i`

Show Worked Solution

`(1 – sqrt 3 i) \ \ => \ r_1=sqrt (1^2 + (sqrt 3)^2)=2`

`theta_1 = tan^(-1) (-sqrt 3)=- pi/3`

`(1 + sqrt 3 i) \ \ => \ r_2=2, \ theta_2 =pi/3`

`:. (1 – sqrt 3 i)^4/(1 + sqrt 3 i)`	`= (2 text(cis) (-pi/3))^4/(2 text(cis) (pi/3))`
	`= (16 text(cis) (-(4 pi)/3))/(2 text(cis) (pi/3))`
	`= 8 text(cis) (-(5 pi)/3)`
	`= 8 text(cis) (pi/3)\ \ \ (-pi<=theta<=pi)`
	`= 8(1/2 + sqrt 3/2 i)`
	`= 4 + 4 sqrt 3 i`

Complex Numbers, SPEC2 2015 VCAA 7 MC

If `z = sqrt3 + 3i`, then `z^63` is

real and negative
equal to a negative real multiple of `i`
real and positive
equal to a positive real multiple of `i`
a positive real multiple of `1 + isqrt3`

Show Answers Only

`A`

Show Worked Solution

`z`	`= sqrt3 + 3i`
	`= 2sqrt3 text(cis)\ (pi/3)`

`:. z^63`	`= (2sqrt3)^63 text(cis)\ ((63pi)/12)`
	`= (2sqrt3)^63 text(cis)\ (21pi)`
	`= (2sqrt3)^63 text(cis)\ (pi)`
	`= -(2sqrt3)^63`

`=> A`

Complex Numbers, SPEC2 2015 VCAA 5 MC

Given `z = (1 + isqrt3)/(1 + i)`, the modulus and argument of the complex number `z^5` are respectively

`2sqrt2` and `(5pi)/6`
`4sqrt2` and `(5pi)/12`
`4sqrt2` and `(7pi)/12`
`2sqrt2` and `(5pi)/12`
`4sqrt2` and `-pi/12`

Show Answers Only

`B`

Show Worked Solution

`z_1 = 1 + isqrt3`

`r`	`= sqrt(1 + 3)=2`
`theta`	`= tan^(−1)(sqrt3)= pi/3`

`z_1 = 2\ text(cis)(pi/3)`

`z_2 = 1 + i`

`r`	`= sqrt(1 + 1)=sqrt2`
`theta`	`= tan^(−1)(1)=pi/4`

`z_1/z_2`	`= 2/sqrt2\ text(cis)(pi/3 – pi/4)`
	`= sqrt2\ text(cis)(pi/12)`

`:. z^5`	`= (sqrt2)^5\ text(cis)((5pi)/12)\ \ \ text{(De Moivre)}`
	`= 4sqrt2\ text(cis)((5pi)/12)`

`=> B`

Complex Numbers, SPEC2 2014 VCAA 8 MC

The principal argument of `(−3sqrt2 - isqrt6)/(2 + 2i)` is

A. `(−13pi)/12`

B. `(7pi)/12`

C. `(11pi)/12`

D. `(13pi)/12`

E. `(−11pi)/12`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ z_1=−3sqrt2 – isqrt6`

`=>\ text(Arg)\ z_1 = pi + tan^(-1) (sqrt6/(3 sqrt2)) = (7pi)/6`

`text(Let)\ \ z_2 = 2 + 2i`

`=>\ text(Arg)\ z_2 = pi/4`

`:.\ text(Arg)\ (z_1/z_2)`	`= text(Arg)\ ((−3sqrt2 – isqrt6)/(2 + 2i))`
	`= (7pi)/6 – pi/4`
	`= (11pi)/12`

`=> C`

Complex Numbers, SPEC2 2018 VCAA 5 MC

Let `z = a + bi`, where `a, b in R\ text(\ {0})`

If `z + 1/z \ in R`, which one of the following must be true?

`text(Arg)(z) = pi/4`
`a = -b`
`a = b`
`|z| = 1`
`z^2 = 1`

Show Answers Only

`D`

Show Worked Solution

`z + 1/z`	`= a + bi + 1/(a + bi)`
	`= a + bi + (a – bi)/{(a + bi)(a – bi)}`
	`= a + bi + (a – bi)/(a^2 + b^2)`
	`= {(a + bi)(a^2 + b^2) + a – bi}/(a^2 + b^2)`
	`= {a(a^2 + b^2) + a}/(a^2 + b^2) + {b (a^2 + b^2) – b}/(a^2 + b^2)\ i`

`text(If)\ \ z + 1/z in R\ \ =>\ {b (a^2 + b^2) – b}/(a^2 + b^2) = 0`

`b(a^2 + b^2) – b = 0`

`b(a^2 + b^2 – 1) = 0`

`a^2 + b^2 = 1,\ \ (b !=0)`

`a^2 + b^2 = |z|^2 = 1`

`:. |z| = 1`

`=> D`

Complex Numbers, SPEC1 2018 VCAA 2

Show that `1 + i = sqrt 2\ text(cis)(pi/4)`. (1 mark)
Evaluate `(sqrt 3 - i)^10/(1 + i)^12`, giving your answer in the form `a + bi`, where `a, b ∈ R`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`-8 – 8 sqrt 3 i`

Show Worked Solution

a.	`r`	`= sqrt(1^2 + 1^2)`
		`= sqrt 2`
	`theta`	`= tan^(-1) (1/1) = pi/4`

	`:. 1 + i`	`= sqrt 2\ text(cis)(pi/4)`

b.	`r_2`	`= sqrt((sqrt 3)^2 + (-1)^2)`
		`= sqrt (3 + 1)`
		`= 2`

`theta_2 = -tan^(-1) (1/sqrt 3)=-pi/6`

`sqrt 3 – i`	`= 2\ text(cis)(- pi/6)`
`(sqrt 3 – i)^10`	`= 2^10\ text(cis) ((-10 pi)/6)`

`=>(1 + i)^12`	`= (sqrt 2)^12text(cis)((12pi)/4)`
	`=2^6 text(cis)(3pi)`

`:. (sqrt 3 – i)^10/(1 + i)^12`	`= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
	`= 2^4 text(cis)((-5pi)/3 – (9pi)/3)`
	`= 16 text(cis)((-14pi)/3)`
	`= 16 text(cis) ((-2pi)/3)`
	`= 16(-1/2 + (-sqrt3)/2 i)`
	`= -8 – 8 sqrt 3 i`

Complex Numbers, SPEC1 2011 VCAA 4

Consider `z = (1 - sqrt 3 i)/(-1 + i),\ \ z in C.`

Find the principal argument of `z` in the form `k pi,\ k in R.` (3 marks)

Show Answers Only

`text(Arg)(z) = 11/12 pi`

Show Worked Solution

`z_1`	`= 1 – sqrt3 i`
	`= sqrt(1^2 + 3)\ text(cis)(−tan^(−1)(sqrt3))`
	`= 2text(cis)(−pi/3)`

`z_2`	`= −1 + i`
	`= sqrt(1^2 + 1^2)\ text(cis)(pi – tan^(−1)(1))`
	`= sqrt2 text(cis)(pi – pi/4)`
	`= sqrt2 text(cis)((3pi)/4)`

`text(Arg)((1 – sqrt3 i)/(−1 + i))`	`= text(Arg)(1 – sqrt3 i) – text(Arg)(−1 + i)`
	`= -pi/3 – (3pi)/4`
	`= -(13pi)/12`

`-(13pi)/12 + 2pi`	`= (-13pi+ 24pi)/12`
	`= (11pi)/12`