MCQ
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:
- Ai
- B-1
- ✓-i
- D4
✓
Answer
Correct option: C.
-i
Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
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