Imagining the imaginary

Algebra Level 3

A student has a lesson on Complex Numbers and makes the following conclusions:

$\mathbb{R} \subset \mathbb{I}$
$\mathbb{I} \subset \mathbb{C}$
$\sqrt[-1]{-i} = i$
$\sqrt[]{\bar{z}z} = |z|$
$z = r\cos(\theta)+ir\sin(\theta)$

Which of the above is/are false ?

Notes:

$i = \sqrt {-1}$ is the imaginary unit
$\mathbb{R}$ is the set of all real numbers
$\mathbb{I}$ is the set of all imaginary numbers
$\mathbb{C}$ is the set of all complex numbers
$z=a+ ib; a,b \in \mathbb R$ is the general form of a complex number
$\bar{z} = a-ib$ is the conjugate of $z$
$|z| = \sqrt{a^2+b^2}$ is the modulus of $z$

All of them are true 1 2 3 4 5 None of them is true

1 solution

Viki Zeta
Sep 3, 2016

$\text{1. Now, every Real numbers are Imaginary numbers. Take for example, } \\ \sqrt[]{2} = -\sqrt[]{2} * i^2, \text{ which is purely-real but imaginary} \\ \text{2. Every imaginary number is a subset of complex number, so it's true.} \\ \text{3. Consider the following equation} \\ i^{-1} = \dfrac{1}{i} = \dfrac{1}{i} \times \dfrac{i}{i} \\ = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i \\ i^{-1} = -i \\ i^{\dfrac{-1}{-1}} = -i^{\dfrac{1}{-1}} \\ i = \sqrt[-1]{-i} \\ \text{4. Using properties of complex numbers we have, } \\ \dfrac{1}{z} = \dfrac{\bar{z}}{|z|^2} \\ |z|^2 =z\bar{z} \\ |z| = \sqrt[]{z\bar{z}} \\ \text{5. Using Polar form of complex numbers, we can state that.}$

It is not true that "all real numbers are imaginary". Real number are of the form $a + 0 i$ . Imaginary numbers are of the form $0 + bi$ .

$\sqrt{2}$ is real, and $\sqrt{2} i$ is imaginary. The only real and imaginary number is 0.

Calvin Lin Staff - 4 years, 9 months ago

Imagining the imaginary

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