Is this true for complex number: 4Re(z_1 z_2 ^(−) )=∣z_1 +z_2 ^(−) ∣^2 −∣z_1 −z


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 182011 by mathocean1 last updated on 03/Dec/22

$I s t h i s t r u e f o r c o m p l e x n u m b e r :$ $4 R e (z_{1} \overset{―}{z_{2}}) =∣ z_{1} + \overset{―}{z_{2}} ∣^{2} - ∣ z_{1} - \overset{―}{z_{2}} ∣^{2}$
Commented by Frix last updated on 03/Dec/22

$The question is strange . If you only use$ ${\bar{z}}_{2} you could instead use z_{3} because it will$ $make no difference .$ $Anyway as posted {it}^{'} s wrong .$
	Answered by alephzero last updated on 03/Dec/22

	$z_{1} = a + b i$ $z_{2} = c + d i$ ${\bar{z}}_{2} = c - d i$ $z_{1} {\bar{z}}_{2} = (a + b i) (c - d i) =$ $= a (c - d i) + b i (c - d i) =$ $= a c - a d i + b c i + b d =$ $= a c + b d - a d i + b c i =$ $= a c + b d - i (a d + b c)$ $4 Re (z_{1} {\bar{z}}_{2}) = 4 (a c + b d)$ $∣ z_{1} + {\bar{z}}_{2} ∣^{2} - ∣ z_{1} - {\bar{z}}_{2} ∣^{2} =$ $= ∣ a + b i + c - d i ∣^{2} - ∣ a + b i - c + d i ∣^{2} =$ $= ∣ a + c + i (b - d) ∣^{2} - ∣ a - c + i (b + d) ∣^{2} =$ ${(\sqrt{{(a + c)}^{2} + {(b - d)}^{2}})}^{2} - {(\sqrt{{(a - c)}^{2} + {(b + d)}^{2}})}^{2}$ $= {(a + c)}^{2} + {(b - d)}^{2} - {(a - c)}^{2} - {(b + d)}^{2} =$ $= a^{2} + 2 a c + c^{2} + b^{2} - 2 b d + d^{2} - a^{2} + 2 a c - c^{2} - b^{2} - 2 b d - d^{2} =$ $= 2 a c + 2 a c = 4 a c \neq 4 (a c + b d)$ $\Rightarrow t h i s i s f a l s e$
	Commented by alephzero last updated on 03/Dec/22

	$S o r r y, I^{'} v e c o r r e c t e d n o w, t h a n k y o u!$
	Commented by Frix last updated on 03/Dec/22

	$Error of sign :$ $= {(a + c)}^{2} + {(b - d)}^{2} - {(a - c)}^{2} \overset{!}{-} {(b + d)}^{2} =$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum