Question
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:
- i
- -1
- -i
- 4
Solution:
Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
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if f(x) = x, x < 0: f(x) = 0, x = 0; f(x) = x, x > 0, then $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})$ is equal to:
If x < 5, then.
$-\text{x} < – 5 $
$-\text{x}\leq-5$
$-\text{x} > – 5 $
$-\text{x}\leq-5$
Equation of horizontal line above x-axis at 5 units from x-axis is: