Let z\epsilon C  with Im(z)=10  and  it satisfies 

\frac{2z-n}{2z+n}=2i-1  for some natural number n . Then : 

  • Option 1)

     n = 20 and Re(z) = -10

  • Option 2)

     n = 40 and Re(z) = 10

  • Option 3)

     n = 40 and Re(z) = -10

  • Option 4)

     n = 20 and Re(z) = 10

 

Answers (1)

\frac{2z-n}{2z+n}=2i-1 , z\epsilon C , Im(z)=10

\\let\;z=x+10i\\\frac{2(x+10i)-n}{2(x+10i)+n}=2i-1\\2x-n+20i=(2i-1)(2(x+10i)+n)=(2i-1)((2x+n)+20i)\\2x-n+20i=4xi+2ni-40-2x-n-20i\\2x-n+20i=(-40-2x-n)+i(4x+2n-20)\\compare\;real\;and\;img\;part\\2x-n=-40-2x-n\\20=4x+2n-20\\x=-10\\n=40


Option 1)

 n = 20 and Re(z) = -10

Option 2)

 n = 40 and Re(z) = 10

Option 3)

 n = 40 and Re(z) = -10

Option 4)

 n = 20 and Re(z) = 10

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