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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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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\(b=\sqrt{3} \)
\(-\dfrac{\pi}{2} = \arg (z) = \arg[(b-i)^3]\)
\(b-i\ \Rightarrow \ \text{complex number that lives on}\ \ y=-i \)
\(3 \times \arg(b-i) = -\dfrac{\pi}{2} \)
\(\arg(b-i) = -\dfrac{\pi}{6} \)
\(\tan^{-1}\Big{(}\dfrac{-1}{b} \Big{)} \) | \(= -\dfrac{\pi}{6} \) | |
\(-\dfrac{1}{b}\) | \(=-\dfrac{1}{\sqrt{3}}\) | |
\(\therefore b\) | \(=\sqrt{3}\) |
For `z ∈ C`, if `text(Im)(z) > 0`, then `text(Arg)((zbarz)/(z - barz))` is
`A`
`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`
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)`
`{z: text(Arg)(z_1) <= text(Arg)(z) <= text(Arg)(z_1^4)} ∩ {z: 1 <= |\ z\ | <= 2}, z ∈ C`. (2 marks)
ii. `z_1^4 = text(cis) pi/3`
ii. `z_1^n = ±i\ text(for)\ k ∈ Z`
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)` |
c. |
d. `text(Area of large sector)`
`=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`
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):`
`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`
The principal arguments of the solutions to the equation `z^2 = 1 + i` are
`C`
`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`
Write `(1 - sqrt 3 i)^4/(1 + sqrt 3 i)` in the form `a + bi`, where `a` and `b` are real constants. (3 marks)
`4 + 4 sqrt 3 i`
`(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` |
If `z = sqrt3 + 3i`, then `z^63` is
`A`
`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`
Given `z = (1 + isqrt3)/(1 + i)`, the modulus and argument of the complex number `z^5` are respectively
`B`
`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`
The principal argument of `(−3sqrt2 - isqrt6)/(2 + 2i)` is
A. `(−13pi)/12`
B. `(7pi)/12`
C. `(11pi)/12`
D. `(13pi)/12`
E. `(−11pi)/12`
`C`
`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`
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?
`D`
`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`
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` |
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)
`text(Arg)(z) = 11/12 pi`
`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` |