Is-this-true-for-complex-number-4Re-z-1-z-2-z-1-z-2-2-z-1-z-2-2-

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} =

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