Complex Numbers And Quadratic Equations question 825
Question: The square root of 3 - 4i is [RPET 1999]
Options:
A) $ \pm (2+i) $
B) $ \pm (2-i) $
C) $ \pm (1-2i) $
D) $ \pm (1+2i) $
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Answer:
Correct Answer: B
Solution:
Let $ \sqrt{3-4i}=x+iy $
$ \Rightarrow 3-4i=x^{2}-y^{2}+2ixy $
$ \Rightarrow x^{2}-y^{2}=3, $ $ 2xy=-4 $ ……(i)
$ \Rightarrow {{(x^{2}+y^{2})}^{2}}={{(x^{2}-y^{2})}^{2}}+4x^{2}y^{2} $ $ ={{(3)}^{2}}+{{(-4)}^{2}} $ = 25
$ \Rightarrow x^{2}+y^{2}=5 $ …….(ii) From equation (i) and (ii) $ x^{2}=4\Rightarrow x=\pm 2 $ , $ y^{2}=1 $
$ \Rightarrow y=\pm 1. $ Hence the square root of $ (3-4i) $ is $ \pm (2-i) $ .
BETA
