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Question Number 83966 by Rio Michael last updated on 08/Mar/20

prove that for any complex number z, if ∣z∣ < 1, then Re(z + 1) > 0

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{complex}\:\mathrm{number}\:{z},\:\mathrm{if}\: \\ $$ $$\:\mid{z}\mid\:<\:\mathrm{1},\:\mathrm{then}\:\mathrm{Re}\left({z}\:+\:\mathrm{1}\right)\:>\:\mathrm{0} \\ $$

Answered by mr W last updated on 08/Mar/20

let z=a+bi ∣z∣=(√(a^2 +b^2 ))<1 ⇒a^2 +b^2 <1 ⇒a^2 <1−b^2 ≤1 ⇒−1<a<1 ⇒0<a+1<2 Re(z+1)=Re(a+1+bi)=a+1>0

$${let}\:{z}={a}+{bi} \\ $$ $$\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }<\mathrm{1} \\ $$ $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} <\mathrm{1} \\ $$ $$\Rightarrow{a}^{\mathrm{2}} <\mathrm{1}−{b}^{\mathrm{2}} \leqslant\mathrm{1} \\ $$ $$\Rightarrow−\mathrm{1}<{a}<\mathrm{1} \\ $$ $$\Rightarrow\mathrm{0}<{a}+\mathrm{1}<\mathrm{2} \\ $$ $$ \\ $$ $${Re}\left({z}+\mathrm{1}\right)={Re}\left({a}+\mathrm{1}+{bi}\right)={a}+\mathrm{1}>\mathrm{0} \\ $$

Commented byRio Michael last updated on 08/Mar/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by TANMAY PANACEA last updated on 08/Mar/20

z=rcosθ+irsinθ ∣z∣=r<1 z+1 =1+rcosθ+irsinθ ∣z+1∣ =(√((1+rcosθ)^2 +(rsinθ)^2 )) =(√(1+2rcosθ+r^2 )) (√(1+2rcosθ+r^2 )) >(√(1−2r+r^2 )) when cosθ=−1 (√(1+2rcosθ+r^2 )) <(√(1+2r+r^2 )) when cosθ=1 1+r>(√(1+2rcosθ+r^2 )) >1−r>0 note 1−r>0

$${z}={rcos}\theta+{irsin}\theta \\ $$ $$\mid{z}\mid={r}<\mathrm{1} \\ $$ $${z}+\mathrm{1} \\ $$ $$=\mathrm{1}+{rcos}\theta+{irsin}\theta \\ $$ $$\mid{z}+\mathrm{1}\mid \\ $$ $$=\sqrt{\left(\mathrm{1}+{rcos}\theta\right)^{\mathrm{2}} +\left({rsin}\theta\right)^{\mathrm{2}} }\: \\ $$ $$=\sqrt{\mathrm{1}+\mathrm{2}{rcos}\theta+{r}^{\mathrm{2}} }\: \\ $$ $$\sqrt{\mathrm{1}+\mathrm{2}{rcos}\theta+{r}^{\mathrm{2}} }\:>\sqrt{\mathrm{1}−\mathrm{2}{r}+{r}^{\mathrm{2}} }\:\:{when}\:{cos}\theta=−\mathrm{1} \\ $$ $$\sqrt{\mathrm{1}+\mathrm{2}{rcos}\theta+{r}^{\mathrm{2}} }\:<\sqrt{\mathrm{1}+\mathrm{2}{r}+{r}^{\mathrm{2}} }\:\:{when}\:{cos}\theta=\mathrm{1} \\ $$ $$\mathrm{1}+{r}>\sqrt{\mathrm{1}+\mathrm{2}{rcos}\theta+{r}^{\mathrm{2}} }\:\:>\mathrm{1}−{r}>\mathrm{0} \\ $$ $$\boldsymbol{{note}}\:\mathrm{1}−\boldsymbol{{r}}>\mathrm{0} \\ $$ $$ \\ $$

Commented byRio Michael last updated on 08/Mar/20

thanks sir

$${thanks}\:{sir} \\ $$

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