A) Purely imaginary
B) Real and positive
C) Real and negative
D) None of these
Correct Answer: A
Solution :
Let \[{{z}_{1}}=a+ib=(a,b)\]and \[{{z}_{2}}=c-id=(c,-d)\] Where \[a>0\]and \[d>0\] ......(i) Then \[|{{z}_{1}}|=|{{z}_{2}}|\]Þ \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}+{{d}^{2}}\] Now \[\frac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\frac{(a+ib)+(c-id)}{(a+ib)-(c-id)}\] \[=\frac{[(a+c)+i(b-d)][(a-c)-i(b+d)]}{[(a-c)+i(b+d)][(a-c)-i(b+d)]}\] \[=\frac{({{a}^{2}}+{{b}^{2}})-({{c}^{2}}+{{d}^{2}})-2(ad+bc)i}{{{a}^{2}}+{{c}^{2}}-2ac+{{b}^{2}}+{{d}^{2}}+2bd}\] \[\frac{-(ad+bc)i}{{{a}^{2}}+{{b}^{2}}-ac+bd}\] [using (i)] \[\therefore \]\[\frac{({{z}_{1}}+{{z}_{2}})}{({{z}_{1}}-{{z}_{2}})}\] is purely imaginary. However if \[ad+bc=0\], then \[\frac{({{z}_{1}}+{{z}_{2}})}{({{z}_{1}}-{{z}_{2}})}\] will be equal to zero. According to the conditions of the equation, we can have \[ad+bc=0\] Trick: Assume any two complex numbers satisfying both conditions i.e., \[{{z}_{1}}\ne {{z}_{2}}\]and \[|{{z}_{1}}|\,=\,|{{z}_{2}}|\] Let \[{{z}_{1}}=2+i,{{z}_{2}}=1-2i,\]\[\therefore \,\,\frac{{{z}_{1}}+{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\frac{3-i}{1+3i}=-i\] Hence the result.
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