AP EAMCET · Maths · Complex Number
If \(\sqrt{-3-4 i}=\mathrm{re}^{\mathrm{i} \theta}\), then \(\mathrm{r}^2 \tan \theta=\)
- A \(-5\)
- B \(5\)
- C \(10\)
- D \(-10\)
Answer & Solution
Correct Answer
(A) \(-5\)
Step-by-step Solution
Detailed explanation
Given : \(\sqrt{-3-4 i}=r e\) ...(i) Let \(\sqrt{-3-4 i}=x+i y\) ...(ii) \(\Rightarrow \quad-3-4 i=(x+i y)^2=x^2-y^2+2 i x y\) Comparing both sides, \(x^2-y^2=-3\) \(\Rightarrow 2 x y=-4 \Rightarrow x y=-2\) Solving above equations, we get \(x= \pm 1\) and \(y=\mp 2\) So, we…
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