A) 1 or \[i\]
B) \[i\] or \[-i\]
C) 1 or - 1
D) \[i\]or -1
Correct Answer: C
Solution :
Let \[z=a+ib,|z|\le 1\]Þ \[{{a}^{2}}+{{b}^{2}}\le 1\] and \[w=c+id,|w|\,\le 1\]Þ \[{{c}^{2}}+{{d}^{2}}\le 1\] \[|z+iw|\,=\,|a+ib+i(c+id)|=2\] Þ \[{{(a-d)}^{2}}+{{(b+c)}^{2}}=4\] ......(i) \[|z-i\overline{w}|\,=\,|a+ib-i(c-id)|\] Þ \[{{(a-d)}^{2}}+{{(b-c)}^{2}}=4\] ......(ii) From (i) and (ii), we get \[bc=0\] Þ Either \[b=0\]or \[c=0\] If\[b=0\], then\[{{a}^{2}}\le 1\]. Then, only possibility is \[a=1\]or\[-1\].
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