0
$\begingroup$

If $\left | z \right | \leq 1$, which of the following must be true?

Indicate all such statements

A. $z^{2}\leq1$

B. $z^{2}\leq z$

C. $z^{3}\leq z$

Not sure if this is right, but is $\left | z \right | = z^{2}$?

For A), I have $-1\leq z \leq 1$ from $\left | z \right | \leq 1$. But I get $z \leq \pm 1$ from $z^{2}\leq1$.

For B), I get $z \leq 0$ and $z\leq 1$ from $z^{2}\leq z$.

For C), I get $z \leq 0$ and $z \leq \pm 1$ from $z^{3}\leq z$.

I know that the answer is A, but I don't understand why.

  • 0
    is $z$ a real number? – 2017-02-16
  • 0
    @S.C.B. It's a real number. – 2017-02-16
  • 1
    (A) It's $z\leq1$ not $z\leq\pm1$. – 2017-02-16
  • 0
    (B) $z=-\dfrac12$. – 2017-02-16
  • 0
    @MyGlasses Thanks for your input. I am not entirely comfortable working with inequalities, so may I know why? – 2017-02-16
  • 0
    If $x$ is a positive number less than $1$, then $0<\cdots – 2017-02-16
  • 1
    `is |z|=z^2` No, $|z| = \sqrt{z^2}\,$ for $z \in \mathbb{R}$. – 2017-02-16

1 Answers 1

1

$$\begin{eqnarray*} &|z|\le 1 \iff-1 \le z \le 1 \\ \\A.& \;(z+1)(z-1)\le 0 &\iff -1 \le z \le 1 \\B.& \;z(z-1)\le 0 &\iff 0 \le z \le 1 \\C.& \;z(z-1)(z+1)\le 0 &\iff z\le -1 \text{ or } 0 \le z \le 1 \end{eqnarray*}$$

Statement A. is the only one that is true whenever $|z|\le 1$

  • 0
    For B, why is $z \geq 0$? Isn't it $z \leq 0$ or $z \leq 1$? Same type of a problem for me to understand C. – 2017-02-16
  • 1
    The graph of $f(z)=z(z-1)$ is a parabola with roots at $z=0$ and $z=1$, the graph only dips below the axis between the two roots – 2017-02-16