Opening the door
What happens when we add together two complex numbers?
Problem
This resource is part of our Adventures with Complex Numbers collection
In the article Vanishing Roots, we posed the question:
"What if we could have the square root of a negative number?"
and introduced $i$ as a square root of $-1$, so $i^2=-1$.
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A complex number has a real part and an imaginary part, for example:
$4+ 2i$,
or $\frac{7}{5} - 3i$,
or $-\sqrt{3} + \pi i$
The complex number $a + bi$ has real part $a$ and imaginary part $b$.
In the same way that we use the number line to represent real numbers, we can use a coordinate plane to represent complex numbers. This plane is called the Argand diagram.
Adding and subtracting complex numbers is just the same as collecting like terms in algebra. For example,
$(4 + 2i) + \left(\frac{7}{5} - 3i \right) = \frac{27}{5} - i$
and
$(4 + 2i) - \left(\frac{7}{5} - 3i \right) = \frac{13}{5} + 5i$.
We have created the GeoGebra interactivity below for you to explore the questions that follow.
Use the Geogebra interactivity to find some pairs of complex numbers whose sum is a real number. What do you notice?
Can you explain it algebraically?
Use the Geogebra interactivity to find some pairs of complex numbers whose sum is an imaginary number. What do you notice?
Can you explain it algebraically?
In general, what would you need to add to $a + bi$ to get a real number? Or to get an imaginary number?
Now that you've been introduced to the world of complex numbers, you might like to start Strolling Along.
$4+ 2i$,
or $\frac{7}{5} - 3i$,
or $-\sqrt{3} + \pi i$
The complex number $a + bi$ has real part $a$ and imaginary part $b$.
In the same way that we use the number line to represent real numbers, we can use a coordinate plane to represent complex numbers. This plane is called the Argand diagram.
Adding and subtracting complex numbers is just the same as collecting like terms in algebra. For example,
$(4 + 2i) + \left(\frac{7}{5} - 3i \right) = \frac{27}{5} - i$
and
$(4 + 2i) - \left(\frac{7}{5} - 3i \right) = \frac{13}{5} + 5i$.
We have created the GeoGebra interactivity below for you to explore the questions that follow.
Use the Geogebra interactivity to find some pairs of complex numbers whose sum is a real number. What do you notice?
Can you explain it algebraically?
Use the Geogebra interactivity to find some pairs of complex numbers whose sum is an imaginary number. What do you notice?
Can you explain it algebraically?
In general, what would you need to add to $a + bi$ to get a real number? Or to get an imaginary number?
Now that you've been introduced to the world of complex numbers, you might like to start Strolling Along.
Student Solutions
Thank you very much to Pablo and Sergio from King's College Alicante in Spain, Amrit from Hymers College in the UK, Jacob and Michael from Cedar House School in South Africa, Alex from Bexley Grammar School in the UK and Niklas from Walton High School in the UK, who all sent in good work on this problem. The solution below draws on their observations and explanations.
Moving $z_1$ and $z_2$ about, there are lots of ways to make $z_3=z_1+z_2$ a real number, which lies on the real axis.
$z_1$ and $z_2$ need to be on either side of the real axis, and both the same vertical distance from the real axis. They can be directly above each other, but they don't have to be. Here are a couple of examples:
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Notice that moving $z_1$ and $z_2$ horizontally gives other real sums:
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For these last three examples, $z_1=a_1-3i$ and $z_2=a_2+3i$ for some real numbers $a_1$ and $a_2$.
We can check this will always work:
$$z_1+z_2=(a_1-3i)+(a_2+3i)=a_1+a_2-3i+3i=(a_1+a_2)+0i$$
What if we had another pair in the form $a_1+bi$ and $a_2-bi$ (for some real number $b$)?
Then $$z_1+z_2=(a_1+bi)+(a_2-bi)=a_1+a_2+bi-bi=(a_1+a_2)+0i$$ So whenever $z_1=a_1+bi$ and $z_2=a_2-bi$, $z_1+z_2$ will be real.
How it works geometrically
When you add $z_2$ to $z_1$, you add the real parts, $a_1+a_2$, so $z_1$ is moved by $a_2$ along the real axis. The imaginary parts are also added, $b_1+b_2$, so $z_1$ is moved by $b_2$ along the imaginary axis.
This is the same as a translation by the vector from $0+0i$ to $z_2$:
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$(1+2i)+(1-2i)=2+0i$
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$(4+4i)+(-2-4i)=2+0i$
We can also prove algebraically that $z_1+z_2$ is only ever a real number if $z_1=a_1+bi$ and $z_2=a_2-bi$.
Then: $$z_1+z_2=(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$$
This is real only if the imaginary part is equal to $0i$,
so $(b_1+b_2)i=0i$, so $b_1+b_2=0$, which means $b_2=-b_1$.
So $z_1=a_1+bi$ and $z_2=a_2-bi$.
Adding two complex numbers to get an imaginary number is very similar.
Here are some examples:
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Similarly, notice that if $z_1=a+b_1i$ and $z_2=-a+b_2i$ then $z_1+z_2=0+(b_1+b_2)i$, which is imaginary (if $b_1+b_2=0$, then $z_1+z_2=0+0i$, which is real - but $0+0i$ is considered to be both real and imaginary).
Geometrically, now the horizontal distances to the imaginary axis are equal and opposite. This means that translation by $z_2$ 'cancels out' the real component of $z_1$:
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We can prove that $z_1$ and $z_2$ must be of the form $a+b_1i$ and $-a+b_2i$ for their sum to be imaginary just like we proved the similar result for the sum to be real:
Let $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$.
Then: $$z_1+z_2=(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$$
This is imaginary only if the real part is equal to $0$, so $a_1+a_2=0$, so $a_2=-a_1$.
Therefore $z_1=a+b_1i$ and $z_2=-a+b_2i$.