Question

If $z=x+iy,x,y\in R$ and $\left|x\right|+\left|y\right|\le k\left|z\right|$ then $k$ is equal to

• A. $1$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. $\sqrt{4}$

Answer 1

The correct option is B

$\sqrt{2}$

Explanation for the correct option.

Find the value of $k$ satisfying the equation.

It is given that $z=x+iy$, so its magnitude is given as: $\left|z\right|=\sqrt{x^2+y^2}$.

Now, it is known that the square of any real number is non-negative. As, $x,y\in R$ so

$$\begin{gathered} {\left(\left|x\right|-\left|y\right|\right)}^2\ge 0 \\ \Rightarrow {\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|\left|y\right|\ge 0 \\ \Rightarrow {\left|x\right|}^2+{\left|y\right|}^2\ge 2\left|x\right|\left|y\right| \\ \Rightarrow 2\left|x\right|\left|y\right|\le {\left|x\right|}^2+{\left|y\right|}^2 \end{gathered}$$

Now, add ${\left|x\right|}^2+{\left|y\right|}^2$ both sides.

$$\begin{gathered} {\left|x\right|}^2+{\left|y\right|}^2+2\left|x\right|\left|y\right|\le {\left|x\right|}^2+{\left|y\right|}^2+{\left|x\right|}^2+{\left|y\right|}^2 \\ \Rightarrow {\left(\left|x\right|+\left|y\right|\right)}^2\le 2\left({\left|x\right|}^2+{\left|y\right|}^2\right) \\ \Rightarrow {\left(\left|x\right|+\left|y\right|\right)}^2\le 2\left(x^2+y^2\right) \end{gathered}$$

Now, take square root of both sides.

$$\begin{gathered} \sqrt{{\left(\left|x\right|+\left|y\right|\right)}^2}\le \sqrt{2\left(x^2+y^2\right)} \\ \Rightarrow \left|x\right|+\left|y\right|\le \sqrt{2}\sqrt{x^2+y^2} \\ \Rightarrow \left|x\right|+\left|y\right|\le \sqrt{2}\left|z\right|\left[\left|z\right|=\sqrt{x^2+y^2}\right] \end{gathered}$$

So on comparing it with $\left|x\right|+\left|y\right|\le k\left|z\right|$, it is seen that $k$ is equal to $\sqrt{2}$.

Hence, the correct option is B.