Answer to Question #124662 in Geometry for desmond

Question

) If z1 = 1 + 2i, find the set of values of z2 for which
(i) |z1 + z2| = |z1| + |z2| (ii) |z1 + z2| = |z1| − |z2|

Answer 1

Given $z_1 = 1 + 2i$ and let $z_2 = x+iy$ .

(i) We have $|z_1 + z_2| = |z_1| + |z_2|$

$\implies |(1+x)+i(2+y)| = |1+2i|+|x+iy|$

$\implies \sqrt{(1+x)^2+(2+y)^2}= \sqrt{5}+\sqrt{x^2+y^2}$

By squaring on both sides, we get$x^2+y^2+2x+4y+5 = 5 + x^2+y^2 +2\sqrt{5x^2+5y^2}$

$\implies x+2y = \sqrt{5x^2+5y^2}$

Again by squaring on both side, we get $x^2+4y^2 + 4xy = 5x^2+5y^2$

$\implies 4x^2 + y^2 - 4xy = 0$

$\implies(2x-y)^2 = 0 \\ \implies 2x- y = 0 \implies y = 2x$ .

Hence, $z_2 = x(1+2i)$ .

(ii) We have $|z_1 + z_2| = |z_1| - |z_2|$

$\implies |(1+x)+i(2+y)| = |1+2i|-|x+iy|$

$\implies \sqrt{(1+x)^2+(2+y)^2}= \sqrt{5}-\sqrt{x^2+y^2}$

By squaring on both sides, we get

$x^2+y^2+2x+4y+5 = 5 + x^2+y^2 -2\sqrt{5x^2+5y^2}$

$\implies x+2y = -\sqrt{5x^2+5y^2}$

Again by squaring on both side, we get $x^2+4y^2 + 4xy = 5x^2+5y^2$

$\implies 4x^2 + y^2 - 4xy = 0$

$\implies(2x-y)^2 = 0 \\ \implies 2x- y = 0 \implies y = 2x$ .

Hence, $z_2 = x(1+2i)$ .