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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♦♦♦ 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}` |