prove-that-the-absolute-valje-of-z1-z2-lt-absolute-value-of-z1-absolute-value-of-z2-

Question Number 139778 by chiamaka last updated on 01/May/21

p r o v e t h a t t h e a b s o l u t e v a l j e o f z 1 + z 2 <= a b s o l u t e v a l u e o f z 1 + a b s o l u t e v a l u e o f z 2

Answered by mr W last updated on 01/May/21

Commented by mr W last updated on 01/May/21

O A = z_{1}

O B = z_{2}

O C = z_{1} + z_{2}

∣ A C ∣=∣ O B ∣=∣ z_{2} ∣

∣ O C ∣⩽∣ O A ∣ + ∣ A C ∣

\Rightarrow∣ z_{1} + z_{2} ∣⩽∣ z_{1} ∣ + ∣ z_{2} ∣

Answered by mr W last updated on 01/May/21

z_{1} = a + b i

∣ z_{1} ∣= \sqrt{a^{2} + b^{2}}

z_{2} = p + q i

∣ z_{2} ∣= \sqrt{p^{2} + q^{2}}

z_{1} + z_{2} = (a + p) + (b + q) i

∣ z_{1} + z_{2} ∣= \sqrt{{(a + p)}^{2} + {(b + q)}^{2}}

{M i n k o w s k i}^{'} s i n e q u a l i t y :

∣ \sqrt{{(a + p)}^{2} + {(b + q)}^{2}} ⩽ \sqrt{a^{2} + b^{2}} + \sqrt{p^{2} + q^{2}}

i . e .

∣ z_{1} + z_{2} ∣⩽∣ z_{1} ∣ + ∣ z_{2} ∣

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