Complex Numbers and Quadratic Equations — MATHS STD 11 Science — Question

3. ​​​-i

​​​​​​​Solution:

Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$

$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$

$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$

$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$

$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$

$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$

$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$

$\Rightarrow\text{z}=-\text{i}$