Complex Analysis /Chapter 1/

Question

Suppose $z_1$ and $z_2$ are complex numbers. What can be said about $z_1$ or $z_2$ if $z_1 z_2=0$?

This is a question from Dennis G. Zill & Patrick D. Shanahan's book.

I need your help to solve it, I clearly don't understand the question well. I will be thankful.

Answer 1

For any non-zero complex number $z = a + bi$, we have

$\bar z = a - bi, \tag 1$

and

$\vert z \vert = \sqrt{a^2 + b^2} \ne 0; \tag 2$

so,

$\vert z \vert^2 = a^2 + b^2 = (a + bi)(a - bi) = z \bar z; \tag 3$

then

$z \dfrac{\bar z}{\vert z \vert^2} = \dfrac{z \bar z}{\vert z \vert^2} = \dfrac{\vert z \vert^2}{\vert z \vert^2} = 1, \tag 4$

which shows the multiplicative inverse of $z$ is $\bar z / \vert z \vert^2$.

Thus if $z_1 \ne 0$ and

$z_1 z_2 = 0, \tag 5$

then

$z_2 = 1 \cdot z_2 = (\dfrac{\bar z_1}{\vert z_1 \vert^2}z_1) z_2 = \dfrac{\bar z_1}{\vert z_1 \vert^2} (z_1 z_2) = \dfrac{\bar z_1}{\vert z_1 \vert^2} \cdot 0 = 0. \tag 6$

We have thus shown that

$z_1 z_2 = 0 \Longleftrightarrow z_1 = 0 \vee z_2 = 0. \tag 7$

Answer 2

$z_1 = 0$ or $z_2 = 0$

As $\Bbb{C}$ is a field, so it's an integral domain, which has no zero divisor. This implies $z_1 = 0$ or $z_2 = 0$.

Answer 3

If $z_1 * z_2 = 0$ can we say that $z_2 = \frac 0{z_1}$?

Why or why not? Or does it depend on other circumstances? If so what circumstances?

If we can say $z_2 =\frac 0{z_1}$, do we know what $\frac 0{z_1}$ is? Or do we have to know what $z_1$ is first?

If we are still struggling is it possible to solve $z_1 =a + bi$ and $z_2 = c + di$ and $(a+bi)(c + di) = ac +adi + bci - bdi = (ac - bd) + (ad + bc)i = 0 = 0 + 0i$ so $ac - bd = 0$ and $ad + bc = 0$.

That's two equations and four unknowns? Is that solvable?

$ac = bd$ so unless $c$ or $d$ equal $0$ we can have $ac = bd \implies \frac cd = \frac ba$. $ad = -bc$ And so unless $b$ or $d$ equal $0$ we can have $ad=-bc \implies \frac cd = -\frac ab$. So $\frac ba = -\frac ab$ so $b^2 = - a^2$.

So....

Answer 4

Let $z_1=a+bi$ and $z_2=c+di$ with $a,b,c,d\in\mathbb{R}$. Then $$z_1z_2=(a+bi)(c+di)=(ac-bd)+(ad+bc)i=0$$ giving the two equations $$ac-bd=0\implies acd-bd^2=0$$ and $$ad+bc=0\implies acd+bc^2=0$$ Subtraction the first from the second gives $$bc^2-bd^2=b(c-d)(c+d)=0$$

Case $1$: If $b=0$, then $ac=0$ so at least one of $a,c$ has to be $0$, and $ad=0$, so at least one of $a,d$ has to be $0$. Hence we get that either $z_1=a,z_2=0$ or $z_1=z_2=0$.

Case $2$: If $c=d$, then $d(a-b)=0$ and $d(a+b)=0$. This means that either $z_1=a+bi,z_2=0$ or $z_1=z_2=0$.

Case $3$: If $c=-d$, then $d(a+b)=0$ and $d(a-b)=0$. This gives the same results as Case $2$.

Therefore we have that at least one of $z_1,z_2$ is $0$.