A) \[\pm (2+i)\]
B) \[\pm (2-i)\]
C) \[\pm (1-2i)\]
D) \[\pm (1+2i)\]
Correct Answer: A
Solution :
Let \[\sqrt{3-4i}=x+iy\]\[\Rightarrow \,\,3-4i=\,{{x}^{2}}-{{y}^{2}}+2ixy\] \[\Rightarrow {{x}^{2}}-{{y}^{2}}=3,\] \[2xy=-4\] ......(i) \[\Rightarrow \,\,{{({{x}^{2}}+{{y}^{2}})}^{2}}=\,{{({{x}^{2}}-{{y}^{2}})}^{2}}+4{{x}^{2}}{{y}^{2}}\]\[={{(3)}^{2}}+{{(-4)}^{2}}\] = 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 \[\,\pm \,(2-i)\].
You need to login to perform this action.
You will be redirected in
3 sec
