x=2i, y=-2i or vice versa.
Hi MHF
I'm writing my AdMath exam in a few days, so I went through previous years' tests and found a problem that I just can't solve!
It goes like this:
The sum of two complex numbers is 0. The product of these two numbers is 4. What are the two numbers?
At first I thought of doing a simultaneous equation , then I tried using conjugates ... nothing works.
I spent a few hours trying to solve it, using every method I could. Is this a trick question or am I just missing something simple?
Please help!
I don't see why you can't do simultaneous equations...
Let $\displaystyle \begin{align*} z_1 = x_1 + y_1\,\iota \end{align*}$ and $\displaystyle \begin{align*} z_2 = x_2 + y_2\,\iota \end{align*}$, then
$\displaystyle \begin{align*} z_1 + z_2 &= 0 + 0\,\iota \\ x_1 + y_1\,\iota + x_2 + y_2\,\iota &= 0 + 0\,\iota \\ \left( x_1 + x_2 \right) + \left( y_1 + y_2 \right) \, \iota &= 0 + 0\,\iota \\ x_1 + x_2 = 0 \textrm{ and } y_1 + y_2 &= 0 \end{align*}$
Also
$\displaystyle \begin{align*} z_1\,z_2 &= 4 + 0\,\iota \\ \left( x_1 + y_1\,\iota \right) \left( x_2 + y_2 \,\iota \right) &= 4 + 0\,\iota \\ x_1\,x_2 + x_1\,y_2\,\iota + x_2\,y_1\,\iota + y_1\,y_2\,\iota ^2 &= 4 + 0\,\iota \\ x_1\,x_2 - y_1\,y_2 + \left( x_1\,y_2 + x_2\,y_1 \right) \, \iota &= 4 + 0\,\iota \\ x_1\,x_2 - y_1\,y_2 = 4 \textrm{ and } x_1\,y_2 + x_2\,y_1 &= 0 \end{align*}$
So now you need to solve the equations $\displaystyle \begin{align*} x_1 + x_2 = 0, y_1 + y_2 = 0, x_1\,x_2 - y_1\,y_2 = 4, x_1\,y_2 + x_2\,y_1 = 0 \end{align*}$ simultaneously.
Let's see, from the first equation we have $\displaystyle \begin{align*} x_2 = -x_1 \end{align*}$ and from the second we have $\displaystyle \begin{align*} y_2 = -y_1 \end{align*}$. Substituting into the fourth equation gives
$\displaystyle \begin{align*} x_1\,y_2 + x_2\,y_1 &= 0 \\ x_1 \left( -y_1 \right) + \left( -x_1 \right) y_1 &= 0 \\ -2x_1\,y_1 &= 0 \\ x_1 = 0 \textrm{ or } y_1 = 0 \end{align*}$
Case 1: $\displaystyle \begin{align*} x_1 = 0 \implies x_2 = 0 \end{align*}$, thus from the third of the original equations we have
$\displaystyle \begin{align*} x_1\,x_2 -y_1\,y_2 &= 4 \\ y_1 ^2 &= 4 \\ y_1 &= \pm 2 \end{align*}$
and since $\displaystyle \begin{align*} y_2 = -y_1 \end{align*}$ that means $\displaystyle \begin{align*} y_2 = \mp 2 \end{align*}$.
So a possibility is $\displaystyle \begin{align*} z_1 = 2\iota \end{align*}$ and $\displaystyle \begin{align*} z_2 = -2\iota \end{align*}$.
Now try Case 2, with $\displaystyle \begin{align*} y_1 = 0 \end{align*}$.
To me, this seems very simple:
$z_1 + z_2 = 0$ (1)
$z_1z_2 = 4$ (2)
Using $z_2 = -z_1$ from (1) in (2), we get:
$z_1(-z_1) = 4$ so:
$z_1^2 = -4$. or:
$\left(\dfrac{z_1}{2}\right)^2 = -1$.
This means $\dfrac{z_1}{2} = \pm i \implies z_1 = \pm 2i \implies z_2 = \mp 2i$.
Hello, Dude001!
The sum of two complex numbers is 0.
The product of these two numbers is 4.
What are the two numbers?
Their sum is 0.
. . Let be one of the numbers.
. . Then is the other.
Their product is 4: .
. .
Equate real and imaginary components: .
From [2], if , then
From [2], if , then
The two numbers are: .
I find this needlessly complicated, and confusing. Let me point out why.
If we write $z = a+bi$, the tacit assumption is that $a$ and $b$ are REAL. So your "second conclusion" from [2], is unwarranted, there is NO real number $a$ such that $a^2 = -4$.
Secondly, why introduce "more variables"? All that is needed to solve this is the fact that for the complex number $i,\ i^2 = (-i)^2 = -1$ and that these are the ONLY two square roots of the complex number -1.
To repeat, for emphasis: we do not need to "resolve to real numbers" to do algebra in the complex numbers; they form a field, and algebraic operations can be performed on complex numbers DIRECTLY.
It's not that your solution is "incorrect", your answer and ProveIt's answer, and mine all give the same result. It's just that they both introduce various "cases" and several equations which add more computation that does not need to be done.
To throw in with the rest, here is how I would do it. If the sum of two numbers is $a$ and their product is $b$, then the two numbers are the solutions to the polynomial $z^2-az+b=0$. In the case with $a=0,b=4$, this polynomial is $z^2+4=0$, so $z=\pm 2i$.
Indeed, for:
$x + y = a$
$xy = b$ then if $b \neq 0$:
$x + \dfrac{b}{x} = a$
$x^2 + b = ax$
$x^2 - ax + b = 0$.
$b = 0$ is somewhat of a bother: if one of $x,y \neq 0$, we can still use the same method, if both are 0, it follows (from $x = y$) that: $x^2 = 0$.
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A geometric solution:
since $z_1 + z_2 = 0$, these two vectors are equal in magnitude and point in opposite directions. Since the (complex) product multiplies magnitudes, and adds angles of incidence with the real axis, and the product is the real number 4, each vector must have magnitude 2, and have angles which are negatives of each other (that is, they are symmetric about the real axis). It follows that the vectors must then both lie on the imaginary axis, so they are {2i,-2i}$.