complex theory

taking the points P(1,0),Q(-1,0) and Z we see that mod(1+z)=mod(-1-z)=magnitude of vector QZ. similarly mod(1-z)=magnitude of vector PZ. BY TRIANGULAR INEQUALITY, modPZ+modQZ>=modPQ=2 Hence the answer..

$Let's\quad assume\quad that\quad z=x+yi,\quad where\quad i(iota)=\sqrt { -1 } \\ and\quad x\quad and\quad y\quad are\quad real\quad numbers.\\ Now,\quad |1+z|+|1-z|=\quad |(1+x)+(yi)|\quad +\quad |(1-x)-(yi)|\\ \qquad \qquad \qquad \qquad \quad \quad =\quad { (1+x) }^{ 2 }+{ (yi) }^{ 2 }\quad +\quad { (1-x) }^{ 2 }+({ -yi })^{ 2 }\\ \qquad \qquad \qquad \qquad \quad \quad =\quad { x }^{ 2 }+1-{ y }^{ 2 }+2xy+{ x }^{ 2 }+1-2xy-{ y }^{ 2 }\\ Cancelling,\quad |1+z|+|1-z|\quad =\quad 2+2({ x }^{ 2 }-{ y }^{ 2 })\\ Since\quad both\quad x\quad and\quad y\quad are\quad real\quad numbers,\\ So,\quad the\quad minimum\quad value\quad of\quad the\quad difference\quad of\quad the\\ squares\quad of\quad two\quad real\quad numbers\quad is\quad 0.\quad Therefore,\\ { \{ |1+z|+|1-z|\} }_{ min }=\quad 2+2(0)\\ or,\quad { \{ |1+z|+|1-z|\} }_{ min }=\quad 2$

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The solution of this question is the sum of minimum distance of any arbitrary point z from (-1,0) and (1,0) when plotted in the Argand plane...By plotting it we find that the minimum distance will be when z is at the origin...hence min(modulus(1+z)) =1 also min(modulus(1-z))=1..