A) Any real number
B) Any complex number
C) Any natural number
D) None of these
Correct Answer: A
Solution :
Let \[z=x+iy\] ......(i) Given \[|z+i|\,=\,|z-i|\] or \[|x+iy+i|\,=\,|x+iy-i|\] or \[|x+i(y+1)|\,=\,|x+i(y-1)|\] or \[\sqrt{{{x}^{2}}+\,{{(y+1)}^{2}}}=\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}\] or \[{{x}^{2}}+{{(y+1)}^{2}}={{x}^{2}}+{{(y-1)}^{2}}\] or \[{{y}^{2}}+2y+1={{y}^{2}}-2y+1\]or \[4y=0\]or \[y=0\] Hence from (i), we get\[z=x\], where \[x\] is any real number.You need to login to perform this action.
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