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ISI2016-MMA-2

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Let Z = x+iy

given |z+1| = |z+i| .....................@1

          |z| =5 ..............................@2

|z+1| = |a+ib+1| =|a+1 +ib| = √ ((a+1)2+b2)

|z+i|= |a+ib-i| = |a+ i(b-1) |    = √(a2+(b+1)2)

√ (a2+1+2a +b2) = √ (a2 +b2 +2b +1) 

          a=b (squaring both sides and cancelling common parts)...........@3

now using  @2 ....

|z| = √ (x2+y2) = 5             => √2(x)2=5  (as x= y ) 

=> x = +√5 or -√5      and   y = +√5 or -√5 

therefore  values for z = +√5  + i √5          , -√5  + i √5

                                        +√5  - i √5         ,-√5  - i √5 

so required values are  +√5  + i √5     and  -√5  - i √5  (from @2 as x=y)

hence solution = 2 option C.

 

 

 

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