A) \[1\pm 3i\]
B) \[\pm (1-3i)\]
C) \[\pm (1+3i)\]
D) \[\pm (3-i)\]
Correct Answer: B
Solution :
Given that \[\sqrt{-8-6i}=x+iy=z\] Þ \[-8-6i={{(x+iy)}^{2}}\] \[\therefore {{x}^{2}}-{{y}^{2}}=-8\] .....(i) and \[2xy=-6\] .....(ii) Now \[{{x}^{2}}+{{y}^{2}}=\sqrt{64+36}=\pm 10\] .....(iii) From (i) and (iii), we get \[x=\pm 1\]and \[y=\pm 3\] Hence \[z=\pm (1-3i)\] Trick: Since \[{{\{\pm (1-3i)\}}^{2}}=-8-6i\]
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