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

prove that the absolute valje of z1+z2<=absolute value of z1+absolute value of z2

$${prove}\:{that}\:{the}\:{absolute}\:{valje}\:{of}\:{z}\mathrm{1}+{z}\mathrm{2}<={absolute}\:{value}\:{of}\:{z}\mathrm{1}+{absolute}\:{value}\:{of}\:{z}\mathrm{2} \\ $$

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

Commented bymr W last updated on 01/May/21

OA=z_1   OB=z_2   OC=z_1 +z_2   ∣AC∣=∣OB∣=∣z_2 ∣  ∣OC∣≤∣OA∣+∣AC∣  ⇒∣z_1 +z_2 ∣≤∣z_1 ∣+∣z_2 ∣

$${OA}={z}_{\mathrm{1}} \\ $$ $${OB}={z}_{\mathrm{2}} \\ $$ $${OC}={z}_{\mathrm{1}} +{z}_{\mathrm{2}} \\ $$ $$\mid{AC}\mid=\mid{OB}\mid=\mid{z}_{\mathrm{2}} \mid \\ $$ $$\mid{OC}\mid\leqslant\mid{OA}\mid+\mid{AC}\mid \\ $$ $$\Rightarrow\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\leqslant\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid \\ $$

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

z_1 =a+bi  ∣z_1 ∣=(√(a^2 +b^2 ))  z_2 =p+qi  ∣z_2 ∣=(√(p^2 +q^2 ))  z_1 +z_2 =(a+p)+(b+q)i  ∣z_1 +z_2 ∣=(√((a+p)^2 +(b+q)^2 ))  Minkowski′s inequality:  ∣(√((a+p)^2 +(b+q)^2 ))≤(√(a^2 +b^2 ))+(√(p^2 +q^2 ))  i.e.  ∣z_1 +z_2 ∣≤∣z_1 ∣+∣z_2 ∣

$${z}_{\mathrm{1}} ={a}+{bi} \\ $$ $$\mid{z}_{\mathrm{1}} \mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$ $${z}_{\mathrm{2}} ={p}+{qi} \\ $$ $$\mid{z}_{\mathrm{2}} \mid=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$ $${z}_{\mathrm{1}} +{z}_{\mathrm{2}} =\left({a}+{p}\right)+\left({b}+{q}\right){i} \\ $$ $$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\sqrt{\left({a}+{p}\right)^{\mathrm{2}} +\left({b}+{q}\right)^{\mathrm{2}} } \\ $$ $${Minkowski}'{s}\:{inequality}: \\ $$ $$\mid\sqrt{\left({a}+{p}\right)^{\mathrm{2}} +\left({b}+{q}\right)^{\mathrm{2}} }\leqslant\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$ $${i}.{e}. \\ $$ $$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\leqslant\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid \\ $$

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