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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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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\(\operatorname{Re}\left(\dfrac{1}{1-z}\right) = \dfrac{1}{2}\)
\(z=a+b i\)
\(\abs{z}=a^2+b^2=1\)
\(\dfrac{1}{1-z}=\dfrac{1}{1-(a+bi)}=\dfrac{1}{(1-a)-bi} \times \dfrac{(1-a)+bi}{(1-a)+bi}=\dfrac{(1-a)+bi}{(1-a)^2+b^2}\)
| \(\operatorname{Re}\left(\dfrac{1}{1-z}\right)\) | \(=\dfrac{-a+1}{a^2-2 a+b^2+1}\) |
| \(=\dfrac{-a+1}{1-2 a+1}\) | |
| \(=\dfrac{-(a-1)}{-2(a-1)}\) | |
| \(=\dfrac{1}{2}\) |
Which of the following is equal to \((a+i b)^3\)?
\(C\)
| \((a+i b)^3\) | \(=a^3+3a^2ib+3a(ib)^2+(ib)^3\) | |
| \(=a^3+3a^2ib-3ab^2-ib^3\) | ||
| \(=(a^3-3ab^2)+i(3a^2b-b^3) \) |
\(\Rightarrow C\)
Find `overset5 underset{n=1}∑ (i)^n`. (2 marks)
`i`
| `overset5 underset{n=1}∑ (i)^n` | `= i + i^2 + i^3 + i^4 + i^5` |
| `= i – 1 – i + 1 + i` | |
| `= i` |
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
`A`
| `(y – ix)^14` | `= (−i(x + iy))^14` |
| `= −i^14(x + iy)^14` | |
| `= −(x + iy)^14` | |
| `= −a – ib` |
`=>A`
Let `z` be a complex number such that `z^2 = -i bar z`.
Which of the following is a possible value for `z`?
`C`
`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`
Write `i^{9}` in the form `a + ib` where `a` and `b` are real. (1 mark)
`0 + 1i`
| `i^{9}` | `= i xx i^{8}` |
| `= i` | |
| `= 0 + 1i` |