0
$\begingroup$

She explains how to obtain the multiplicative inverse:

In what follows, z=a+bi and w=c+di are complex numbers with a,b,c,d∈R. Is there a multiplicative inverse of z? If so, what is it?

Note that z≠0 implies that not both a and b are 0. We want to see if we can find real numbers c and d so that zw=1, or equivalently,

(ac−bd)+(ad+bc)i=1.

As the right-hand side is a real number, we need the imaginary part on the left-hand side to be zero:

ad+bc=0.

That leaves

ac−bd=1.

Solving for c and d in terms of a and b, we get

c=a/(a^2+b^2) and d=−b/(a^2+b^2).

Isn't the multiplicative inverse of z 1/z? Why was it important for her to express c and d in terms of a and b?

$\endgroup$
3
  • $\begingroup$ What is $i^2$ in the field $\mathbb{C}$ of complex numbers? $\endgroup$ Sep 14, 2014 at 20:39
  • $\begingroup$ Ha. Yeah okay thank you - makes sense now. Edited the question - so it's just the last part about her expressing c and d in terms of a and b. $\endgroup$
    – Hal
    Sep 14, 2014 at 20:40
  • $\begingroup$ With your kind of reasoning, the inverse of $2$ in $\mathbf Z/(6)$ is $1/2$. That is just symbol-pushing, with no content to it. Indeed, there is no multiplicative inverse for $2$ in $\mathbf Z/(6)$. Writing $1$ divided by a symbol does not automatically make it invertible. $\endgroup$
    – KCd
    Feb 17, 2015 at 17:02

4 Answers 4

2
$\begingroup$

Yes, the multiplicative inverse of $z$ is $1/z$ (for nonzero $z$). But the point is, what exactly is the complex number $1/z$?

Remember, a complex number is usually expressed in the form $$(\textrm{real number}) +(\textrm{real number})i$$ Your instructor is trying to find what the two real numbers (the real and imaginary components of $1/z$) are--that's the $c$ and $d$ in her notation.

For example, it may not be obvious that $\frac{1}{1+i}$ and $\frac12 -\frac12 i$ are the same complex number. These are different representations of the multiplicative inverse of $1+i$.

$\endgroup$
1
$\begingroup$

The imaginary number $i$ is defined to be the square root of $-1$. This means that $i^2=-1$, and therefore $$zw=ac+adi+bci+bdi^2=ac+adi+bci+bd(-1)=(ac−bd)+(ad+bc)i.$$

For a complex number $x=p+qi$, where $p$ and $q$ are real numbers, it is common to write $\text{Re}(x)=p$ for the real part of $x$ and $\text{Im}(x)=q$ for the imaginary part of $x$. For any two complex numbers $x$ and $y$, we have an equality $x=y$ if and only if $\text{Re(x)}=\text{Re(y)}$ and $\text{Im}(x)=\text{Im}(y)$.

For the equation $zw=1$ to be true, there must be equalities $\text{Re}(zw)=\text{Re}(1)$ and $\text{Im}(zw)=\text{Im}(1)$. This means that $$ac-bd=\text{Re}(zw)=\text{Re}(1)=1\hspace{15pt}\text{and}\hspace{15pt}ad+bc=\text{Im}(zw)=\text{Im}(1)=0.$$

It's true that if $z\neq0$ then the multiplicative inverse of $z$ may be written as $z^{-1}=1/z$. It can also be written as $$z^{-1}=\frac a{a^2+b^2}+\frac{-b}{a^2+b^2}i$$ where $a$ and $b$ are real numbers.

Given that $z$ is originally expressed in terms of the real numbers $a$ and $b$, it can be useful to express it's inverse in this way as well. If you multiply $(a+bi)$ and $\bigl(\frac a{a^2+b^2}+\frac{-b}{a^2+b^2}i\bigr)$, the product is indeed equal to $1$.

$\endgroup$
1
$\begingroup$

Yes, the multiplicative inverse of $z$ is $1/z$. Your teacher is finding an explicit value for $1/z$. This both proves that it exists (she has displayed it) and is useful in the future for computation.

$\endgroup$
1
$\begingroup$
  • First, remember that $i^{2}=-1$ , be definition.

  • Second, she wanted to find the real numbers $c$ and $d$ such thath $w=\frac{1}{z}$, and that is why she expressed them in terms of $a$ and $b$ (an explicit expression of $w$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .