prove that ∣a+(√(a^2 −b^2 ))∣ + ∣a − (√(a^2 −b^2 ))∣ = ∣a+b∣ +∣a−b∣ a,b ∈ C


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Question Number 192129 by universe last updated on 08/May/23

$p r o v e t h a t$ $∣ a + \sqrt{a^{2} - b^{2}} ∣ + ∣ a - \sqrt{a^{2} - b^{2}} ∣ = ∣ a + b ∣ + ∣ a - b ∣$ $a, b \in C$
	Answered by deleteduser1 last updated on 08/May/23

	$S q u a r i n g b o t h s i d e s$ $(∣ x ∣^{2} = x \bar{x}; ∣ x y ∣=∣ x ∣∣ y ∣; \overset{_____}{x + y} = \bar{x} + \bar{y})$ ${L H S}^{2} =$ $(a + \sqrt{a^{2} - b^{2}}) (\bar{a} + \overset{}{\sqrt{a^{2} - b^{2}}}) + (a - \sqrt{a^{2} - b^{2}}) (\bar{a} - \overset{}{\sqrt{a^{2} - b^{2}}})$ $+ 2 ∣ (a + \sqrt{a^{2} - b^{2}}) (a - \sqrt{a^{2} - b^{2}}) = b^{2} ∣$ $= 2 ∣ a ∣^{2} + 2 ∣ a^{2} - b^{2} ∣ + 2 ∣ b ∣^{2}$ ${R H S}^{2} = (a + b) (\bar{a} + \bar{b}) + (a - b) (\bar{a} - \bar{b}) + 2 ∣ a^{2} - b^{2} ∣$ $= 2 ∣ a ∣^{2} + 2 ∣ a^{2} - b^{2} ∣ + 2 ∣ b ∣^{2}$ $S i n c e L H S a n d R H S w e r e b o t h p o s i t i v e b e f o r e$ $s q u a r i n g b o t h s i d e s$ ${L H S}^{2} = {R H S}^{2} \Rightarrow L H S = R H S ◻$
	Commented by York12 last updated on 08/May/23

	$s i r I w a n n a a s k y o u [s e v e r a l q u e s i o n s,$ $I a m a h i g h s c h o o l s t u d e n t$ $a n d I w a n n a a s k a b o u t b o o k s r e c o m m e n d a t i o n d$ $s o s i r t h a t i s m y t e l e g r a m : b e n g u b l e r$
	Commented by deleteduser1 last updated on 09/May/23

	$I {d o n}^{'} t u s e t e l e g r a m .$
	Commented by York12 last updated on 09/May/23

	$s o s i r h o w c a n I r e a c h y o u o u t$
	Answered by universe last updated on 09/May/23
	$(1/2){∣a+b+a−b+2(√((a+b)(a−b)))∣+∣a+b+a−b−2(√((a+b)(a−b)))∣} let a+b = x and a−b = y (1/2){∣((√x))^2 +((√y))^2 +2(√(xy))∣+∣((√x))^2 +((√y))^2 −2(√(xy))∣} (1/2){∣(√x)+(√y) ∣^2 +∣(√x)− (√y) ∣^2 } let (√x) = u and (√y) = v (1/2){(u+v)(u^� +v^� )+(u−v)(u^� −v^� )} (1/2){uu^� +uv^� +vu^� +vv^� +uu^� −uv^� −vu^� +vv^� } uu^� +vv^� ⇒∣u^2 ∣+∣v^2 ∣ ⇒∣((√x))^2 ∣+∣((√y))^2 ∣ ∣x∣+∣y∣ ⇒ ∣a+b∣ + ∣a−b∣$
	$\frac{1}{2} {∣ a + b + a - b + 2 \sqrt{(a + b) (a - b)} ∣ + ∣ a + b + a - b - 2 \sqrt{(a + b) (a - b)} ∣}$ $l e t a + b = x$ $a n d a - b = y$ $\frac{1}{2} {∣ {(\sqrt{x})}^{2} + {(\sqrt{y})}^{2} + 2 \sqrt{x y} ∣ + ∣ {(\sqrt{x})}^{2} + {(\sqrt{y})}^{2} - 2 \sqrt{x y} ∣}$ $\frac{1}{2} {∣ \sqrt{x} + \sqrt{y} ∣^{2} + ∣ \sqrt{x} - \sqrt{y} ∣^{2}}$ $l e t \sqrt{x} = u a n d \sqrt{y} = v$ $\frac{1}{2} {(u + v) (\bar{u} + \bar{v}) + (u - v) (\bar{u} - \bar{v})}$ $\frac{1}{2} {u \bar{u} + u \bar{v} + v \bar{u} + v \bar{v} + u \bar{u} - u \bar{v} - v \bar{u} + v \bar{v}}$ $u \bar{u} + v \bar{v} \Rightarrow∣ u^{2} ∣ + ∣ v^{2} ∣ \Rightarrow∣ {(\sqrt{x})}^{2} ∣ + ∣ {(\sqrt{y})}^{2} ∣$ $∣ x ∣ + ∣ y ∣ \Rightarrow ∣ a + b ∣ + ∣ a - b ∣$
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