MCQ
If $\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy},$ then possible value of $\sqrt{\text{a}-\text{ib}}$ is:
- A$\text{x}^2+\text{y}^2$
- B$\sqrt{\text{x}^2+\text{y}^2}$
- C$\text{x}+\text{iy}$
- ✓$\text{x}-\text{iy}$
✓
Answer
Correct option: D.
$\text{x}-\text{iy}$
$\sqrt{\text{a}+\text{ib}}=\text{x}+\text{iy}$
Squaring on both the sides, we get,
$\text{a}+\text{ib}=\text{x}^2+(\text{iy})^2+2\text{ixy}$
$\Rightarrow\text{a}+\text{ib}=(\text{x}^2-\text{y}^2)+2\text{ixy}$
$\therefore\text{a}=(\text{x}^2-\text{y}^2)$
and $\text{b}=2\text{xy}$
$\therefore\text{a}-\text{ib}=(\text{x}^2-\text{y}^2)-2\text{ixy}$
$\Rightarrow\text{a}-\text{ib}=\text{x}^2+\text{i}^2\text{y}^2-2\text{ixy} \ [\because\text{i}^2=-1]$
Taking square root on both the sides, we get:
$\sqrt{\text{a}-\text{ib}}=\text{x}-\text{iy}$
Squaring on both the sides, we get,
$\text{a}+\text{ib}=\text{x}^2+(\text{iy})^2+2\text{ixy}$
$\Rightarrow\text{a}+\text{ib}=(\text{x}^2-\text{y}^2)+2\text{ixy}$
$\therefore\text{a}=(\text{x}^2-\text{y}^2)$
and $\text{b}=2\text{xy}$
$\therefore\text{a}-\text{ib}=(\text{x}^2-\text{y}^2)-2\text{ixy}$
$\Rightarrow\text{a}-\text{ib}=\text{x}^2+\text{i}^2\text{y}^2-2\text{ixy} \ [\because\text{i}^2=-1]$
Taking square root on both the sides, we get:
$\sqrt{\text{a}-\text{ib}}=\text{x}-\text{iy}$
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