prove-that-for-any-complex-number-z-if-z-lt-1-then-Re-z-1-gt-0-

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

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

l e t z = a + b i

∣ z ∣= \sqrt{a^{2} + b^{2}} < 1

\Rightarrow a^{2} + b^{2} < 1

\Rightarrow a^{2} < 1 - b^{2} ⩽ 1

\Rightarrow - 1 < a < 1

\Rightarrow 0 < a + 1 < 2

R e (z + 1) = R e (a + 1 + b i) = a + 1 > 0

Commented by Rio Michael last updated on 08/Mar/20

t h a n k y o u s i r

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

z = r c o s θ + i r s i n θ

∣ z ∣= r < 1

z + 1

= 1 + r c o s θ + i r s i n θ

∣ z + 1 ∣

= \sqrt{{(1 + r c o s θ)}^{2} + {(r s i n θ)}^{2}}

= \sqrt{1 + 2 r c o s θ + r^{2}}

\sqrt{1 + 2 r c o s θ + r^{2}} > \sqrt{1 - 2 r + r^{2}} w h e n c o s θ = - 1

\sqrt{1 + 2 r c o s θ + r^{2}} < \sqrt{1 + 2 r + r^{2}} w h e n c o s θ = 1

1 + r > \sqrt{1 + 2 r c o s θ + r^{2}} > 1 - r > 0

n o t e 1 - r > 0

Commented by Rio Michael last updated on 08/Mar/20

t h a n k s s i r