prove-that-z-gt-Re-z-Im-z-2-z-C-

Question Number 192126 by universe last updated on 08/May/23

p r o v e t h a t

∣ z ∣ > \frac{∣ R e (z) ∣ + ∣ I m (z) ∣}{2}, \forall z \in C

Commented by York12 last updated on 09/May/23

s i r h o w c a n I r e a c h y o u o u t, I n e e d t o a s k s e v e r a l q u e s t i o n s

Answered by AST last updated on 08/May/23

L e t z = x + y i; R e (z) = \frac{z + \bar{z}}{2}; I m (z) = \frac{z - \bar{z}}{2 i} = \frac{(z - \bar{z}) i}{- 2}

\Rightarrow∣ R e (z) ∣=∣ x ∣; ∣ I m (z) ∣= \frac{∣ z - \bar{z} ∣}{2} =∣ y ∣; ∣ z ∣= \sqrt{x^{2} + y^{2}}

R e q u i r e d t o p r o v e \sqrt{x^{2} + y^{2}} > \frac{∣ x ∣ + ∣ y ∣}{2}

\equiv x^{2} + y^{2} > \frac{∣ x ∣^{2} + ∣ y ∣^{2} + 2 ∣ x ∣∣ y ∣}{4} \equiv 3 (x^{2} + y^{2}) > 2 ∣ x ∣∣ y ∣

I t i s k n o w n t h a t x^{2} + y^{2} ⩾ 2 \sqrt{x^{2} y^{2}} = 2 ∣ x ∣∣ y ∣

H e n c e, 3 (x^{2} + y^{2}) > 2 ∣ x ∣∣ y ∣\Leftrightarrow∣ z ∣> \frac{∣ R e (z) ∣ + ∣ I m (z) ∣}{2}

Answered by mehdee42 last updated on 08/May/23

l e t : z = a + i b

∥ z ∥= \sqrt{a^{2} + b^{2}} ⩾ \sqrt{a^{2}} =∣ a ∣ (i)

∥ z ∥= \sqrt{a^{2} + b^{2}} ⩾ \sqrt{b^{2}} =∣ b ∣ (i i)

(i) + (i i) \Rightarrow 2 ∥ z ∥>∣ a ∣ + ∣ b ∣ \Rightarrow∥ z ∥> \frac{∣ a ∣ + ∣ b ∣}{2} ✓

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