Question

If $z=a+ib$ where $a>0,b>0$, then

• A. $\left|z\right|\ge \frac{1}{\sqrt{2}}(a-b)$
• B. $\left|z\right|\ge \frac{1}{\sqrt{2}}(a+b)$
• C. $\left|z\right|<\frac{1}{\sqrt{2}}(a+b)$
• D. None of these

Answer 1

The correct option is B $\left|z\right|\ge \frac{1}{\sqrt{2}}(a+b)$

As ${(a-b)}^2\ge 0,a^2+b^2\ge 2ab$ ...(1)

But $\left|z\right|=\sqrt{a^2+b^2};$ si from (1), ${\left|z\right|}^2\ge 2ab$

$\therefore {\left|z\right|}^2+a^2+b^2\ge a^2+b^2+2ab\Rightarrow {\left|z\right|}^2+{\left|z\right|}^2\ge {(a+b)}^2\Rightarrow 2{\left|z\right|}^2\ge {(a+b)}^2$

$\Rightarrow \sqrt{2}\left|z\right|\ge a+b$ as $\left|z\right|$ is positive

$\left|z\right|\ge \frac{1}{\sqrt{2}}(a+b)$