z and w ∈ C proof ∣∣z∣−∣w∣∣ ≤ ∣z−w∣ and ∣z∣−∣w∣≤ ∣z+w∣


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Question Number 33096 by 7991 last updated on 10/Apr/18

$z a n d w \in C$ $p r o o f ∣∣ z ∣ - ∣ w ∣∣ ⩽ ∣ z - w ∣ a n d ∣ z ∣ - ∣ w ∣⩽ ∣ z + w ∣$
	Answered by MJS last updated on 10/Apr/18

	$z = x + y i$ $w = u + v i$ $1 .$ $∣ \sqrt{x^{2} + y^{2}} - \sqrt{u^{2} + v^{2}} ∣⩽ \sqrt{{(x - u)}^{2} + {(y - v)}^{2}}$ ${(\sqrt{x^{2} + y^{2}} - \sqrt{u^{2} + v^{2}})}^{2} ⩽ {(x - u)}^{2} + {(y - v)}^{2}$ $(x^{2} + y^{2}) - 2 \sqrt{x^{2} + y^{2}} \sqrt{u^{2} + v^{2}} + (u^{2} + v^{2}) ⩽ x^{2} - 2 x u + u^{2} + y^{2} - 2 y v + v^{2}$ $- 2 \sqrt{x^{2} + y^{2}} \sqrt{u^{2} + v^{2}} ⩽ - 2 (x u + y v)$ $\sqrt{x^{2} + y^{2}} \sqrt{u^{2} + v^{2}} ⩾ x u + y v$ $if (x u + y v) ⩽ 0 this is true$ $if (x u + y v) > 0 :$ $(x^{2} + y^{2}) (u^{2} + v^{2}) ⩾ {(x u + y v)}^{2}$ $x^{2} u^{2} + x^{2} v^{2} + y^{2} u^{2} + y^{2} v^{2} ⩾ x^{2} u^{2} + 2 x u y v + y^{2} v^{2}$ $x^{2} v^{2} - 2 x u y v + y^{2} u^{2} ⩾ 0$ ${(x v - y u)}^{2} ⩾ 0$ $true$ $2 .$ $if ∣ z ∣ - ∣ w ∣⩽ 0 {it}^{'} s already done$ $if ∣ z ∣ - ∣ w ∣> 0 go on like above$
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