MCQ
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:
- Ai
- B-1
- C-i
- D4
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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The value of $\sin\frac{\pi}{18}+\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}+\sin\frac{5\pi}{18}$ is given by:
$\sin\frac{7\pi}{18}+\sin\frac{4\pi}{9}$
$1$
$\cos\frac{\pi}{6}+\cos\frac{3\pi}{7}$
$\cos\frac{\pi}{9}+\sin\frac{\pi}{9}$
The middle term in the expansion of $\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}$ is:
${^\text{2n}}\text{C}_{\text{n}}$
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
${^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
None of these.