Let `z = sqrt3 - 3 i`
- Express `z` in modulus-argument form. (2 marks)
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- Find the smallest integer `n`, such that `z^n + (overset_z)^n = 0`. (3 marks)
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Let `z = sqrt3 - 3 i`
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i. | `z` | `= sqrt3 – 3 i` |
`|z|` | `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3` |
`tan theta` | `= frac{3}{sqrt3}=sqrt3` |
`theta` | `= frac{pi}{3}` |
`text{arg} (z)` | `= – frac{pi}{3}` |
`therefore z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`
ii. `z^n + (overset_z)^n = 0`
`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3}) – i sin (frac{-pi}{3}) ]^n = 0`
`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3}) – i sin (frac{-n pi}{3}) = 0`
`2 \ cos (frac{-n pi}{3})` | `= 0` |
`cos (frac{n pi}{3})` | `= 0` |
`frac{n pi}{3}` | `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …` |
`frac{n}{3}` | `= frac{(2k + 1)}{2}` |
`n` | `= frac{3 (2k + 1)}{2}` |
`text{Numerator will always be odd ⇒ no solution exists}`
Let `beta = 1-i sqrt3`.
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i. `beta = 1 – i sqrt3`
`| beta | = sqrt(1^2 + (sqrt3)^2) = 2`
`tan theta` | `= frac{sqrt3}{1} = sqrt3` |
`theta` | `= frac{pi}{3}` |
`text{arg} (beta)` | `= -frac{pi}{3}` |
`therefore \ beta = 2 \ text{cis} (-frac{pi}{3})`
ii. | `beta^5` | `= 2^5 \ text{cis} (-frac{pi}{3} xx5)` |
`= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)` | ||
`= 32 \ text{cis} (frac{pi}{3})` |
iii. | `beta^5` | `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )` |
`= 32 ( frac{1}{2} + i frac{sqrt3}{2})` | ||
`= 16 + i 16 sqrt3` |
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i. | `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} )` |
ii. `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}) )` |
iii. `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}` |
iv. | `(frac{1 + i sqrt2}{1 + i})^12` | `= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))` |
`= 64 (cos pi + i sin pi)` | ||
`= – 64` |
Let `z = -1 + i sqrt 3`.
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i. | `|\ z\ |` | `= -1 + i sqrt 3` |
`= sqrt((-1)^2 + (sqrt 3)^2)` | ||
`= 2` |
`tan theta` | `= -sqrt 3` | |
`text(arg)(z)` | `= (2 pi)/3` | |
`:. z` | `= 2 text(cis) (2 pi)/3` |
ii. `z^3 = 2^3 [cos(3 xx (2 pi)/3) + i sin (3 xx (2 pi)/3)]\ \ \ text{(by De Moivre)}`
`= 8(cos 2 pi + i sin 2 pi)`
`= 8(1 + 0i)`
`= 8 + 0i`
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.
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i. |
`|\ 1+i\ |` | `=sqrt(1^2+1^2)=sqrt2` |
`text(arg)(1+i)` | `=pi/4` |
`:. 1 + i =` | `sqrt 2 (cos pi/4 + i sin pi/4)` |
ii. `(1 + i)^17` | `=(sqrt 2)^17 (cos\ pi/4 + i sin\ pi/4)^17` |
`=2^8 sqrt 2 (cos (17 pi)/4 + i sin (17 pi)/4)\ \ \ \ text{(De Moivre)}` | |
`=2^8 sqrt 2 (cos pi/4 + i sin pi/4)` | |
`=2^8 sqrt2(1/sqrt2 + 1/sqrt2 i)` | |
`=2^8 (1 + i)` | |
`=256 + 256 i` |
Given that `z = 1 − i`, which expression is equal to `z^3 ?`
(A) `sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
(B) `2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
(C) `sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
(D) `2 sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
`B`
Given `z = 2(cos\ pi/3 + i sin\ pi/3)`, which expression is equal to `(bar {:z:})^(−1)`?
(A) `1/2(cos\ pi/3 − i sin\ pi/3)`
(B) `2(cos\ pi/3 − i sin\ pi/3)`
(C) `1/2(cos\ pi/3 + i sin\ pi/3)`
(D) `2(cos\ pi/3 + i sin\ pi/3)`
`C`
`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`
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i. | `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)` |
ii. | `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` |
Let `z = 2- i sqrt 3` and `w = 1 + i sqrt 3.`
i. `z = 2 – i sqrt 3\ ,\ \ w = 1 + i sqrt 3`
`bar w = 1 – i sqrt 3`
`z + bar w` | `= 2 – i sqrt 3 + 1 – i sqrt 3` |
`= 3 – i\ 2 sqrt 3` |
ii. `|\ w\ |` | `=sqrt(1^2 + (sqrt3)^2)=2` |
`:.w` | `= 2 (1/2 + i sqrt 3/2)` |
`=2(cos\ pi/3 + i sin\ pi/3)` | |
`= 2 text(cis) pi/3` |
iii. `w^24` | `= 2^24 text(cis)\ (24 xx pi/3)` |
`= 2^24\ text(cis) \ 8 pi` | |
`= 2^24` |