How to Find Distance and Midpoint of Complex Numbers?

We all know how to calculate the distance between two spots and the midpoint between two points in a Cartesian plane, but what happens when we’re dealing with something more complicated like a complex plane?

It turns out that the formulae used to get the distance between two complex numbers and the midpoint between two complex numbers are very similar to the formulae used to determine the distance between two Cartesian points.

Complex numbers are discussed in this part, and we will learn how to use both the midpoint formula and the distance formula for them.

Related Topics

• How to Find the Distance of Two Points
• How to Find Midpoint
• How to Solve the Complex Plane

A step-by-step guide to finding the distance and midpoint of complex numbers

The following is the formula for calculating the distance between two points \((a, b)\) on the complex plane for the complex number \(a + bi\) and \((c, d)\) for the complex number \(c + di\):

\(\color{blue}{d=\sqrt{(c-a)^{2}+(d-b){^2}}}\)

• It does not matter which number you subtract from which; the distance comes out the same. If you subtract the real component of the first integer from the second, you must also subtract the first number’s imaginary part from the second.

Using the midpoint rule from coordinate geometry, we can get the midpoint of any two complex numbers \(z=a+bi\) and \(w=c+di\):

\(\color{blue}{M=\frac{a+c}{2}+(\frac{b+d}{2})i}\)

• Thus, the complex number at the middle of the line is given.

Finding Distance and Midpoint of Complex Numbers, Example 1:

Find the distance between the points \(3+2i\) and \(8+5i\).

To find the distance between complex numbers use this formula: \(\color{blue}{d=\sqrt{(c-a)^{2}+(d-b){^2}}}\).

The distance is: \(d=\sqrt{(8-3)^2+(5-2)^2}=\sqrt{(5)^2+(3)^2}\)

Then: \(i^2=-1\)

\(d=\sqrt{(5)^2+(3)^2}=\sqrt{25+9}=\sqrt{34}\approx 5.83\)

Finding Distance and Midpoint of Complex Numbers, Example 2:

Find the midpoint between the points \(6-4i\) and \(2+2i\)

To find the midpoint between complex numbers use this formula: \(\color{blue}{M=\frac{a+c}{2}+(\frac{b+d}{2})i}\)

\(M= \frac {6+2}{2}+(\frac {-4+2}{2})i\)\(=\)\( \frac {8}{2}+(\frac {-2}{2})i=4-i\)

Exercises for Finding Distance and Midpoint of Complex Numbers

Find the distance and midpoint of complex numbers.

1. \(\color{blue}{(1+2i)}\), \(\color{blue}{(-2+4i)}\)
2. \(\color{blue}{(-6+2i)}\), \(\color{blue}{(-2+5i)}\)
3. \(\color{blue}{(-7i)}\), \(\color{blue}{(5-4i)}\)
4. \(\color{blue}{(2+i)}\), \(\color{blue}{(6+5i)}\)
5. \(\color{blue}{(-7-3i)}\), \(\color{blue}{(3+5i)}\)

1. \(\color{blue}{d=\sqrt{13}}\), \(\color{blue}{M=-\frac{1}{2}+3i}\)
2. \(\color{blue}{d=5}\), \(\color{blue}{M=-4+\frac{7}{2}i}\)
3. \(\color{blue}{d=\sqrt{34}}\), \(\color{blue}{M=\frac{5}{2}-\frac{11}{2}i}\)
4. \(\color{blue}{d=4\sqrt{2}}\), \(\color{blue}{M=4+3i}\)
5. \(\color{blue}{d=2\sqrt{41}}\), \(\color{blue}{M=-2+i}\)

Recommended EffortlessMath Books

For a workbook that builds complex numbers from the ground up, the Pre-Calculus for Beginners walks through complex arithmetic, the complex plane, and polar form with worked examples. For deeper algebra-side coverage of complex numbers, the Algebra II for Beginners covers the same topics with extra practice.

Pre-Calculus for Beginners 2026 The Ultimate Step by Step Guide to Acing Precalculus

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Common Mistake: Do Not Square the i

The distance formula works on points in the complex plane. The number 8 + 5i is the point (8, 5), so its vertical coordinate is 5, not 5i. When you subtract the imaginary parts, use the plain numbers.

• Right: for 3 + 2i and 8 + 5i, the differences are 8 − 3 = 5 and 5 − 2 = 3. Distance = √(25 + 9) = √34.
• Wrong: writing the second difference as 3i and squaring it to get −9. That gives √(25 − 9) = 4, which is too short.

A quick check: the distance can never be smaller than either difference on its own. Here √34 ≈ 5.83 is bigger than both 5 and 3, so it makes sense. Another way to say this is that the distance is the modulus of the difference: (8 + 5i) − (3 + 2i) = 5 + 3i, and |5 + 3i| = √34.

Finding a Missing Endpoint from the Midpoint

Sometimes you know the midpoint and one endpoint. Since M = (z₁ + z₂)/2, you can solve for the other endpoint: z₂ = 2M − z₁.

Example: The midpoint of z₁ and z₂ is 1 + 4i, and z₁ = 3 + 2i. Find z₂.

1. Double the midpoint: 2(1 + 4i) = 2 + 8i.
2. Subtract z₁: (2 + 8i) − (3 + 2i) = −1 + 6i.
3. Check: ((3 + (−1))/2) + ((2 + 6)/2)i = 1 + 4i. It matches.

So z₂ = −1 + 6i.

Frequently Asked Questions

What’s a complex number?

A complex number has the form \(a+bi\), where \(a\) and \(b\) are real and \(i=\sqrt{-1}\). The real part is \(a\); the imaginary part is \(b\). Complex numbers extend the real number system to handle square roots of negative numbers.

How do I plot a complex number?

Use the complex plane (also called the Argand diagram). The horizontal axis is the real axis, the vertical axis is the imaginary axis. Plot \(a+bi\) as the point \((a,b)\). So \(3-2i\) sits 3 units right, 2 units down.

What’s the distance between two complex numbers?

The distance between \(z_1=a_1+b_1 i\) and \(z_2=a_2+b_2 i\) is \(\sqrt{(a_1-a_2)^2+(b_1-b_2)^2}\). This is also written as \(|z_1-z_2|\), the modulus of their difference. It’s the regular distance formula applied to the points \((a_1,b_1)\) and \((a_2,b_2)\) – identical math, just relabeled axes. Example: distance from \(2+3i\) to \(5+7i\) is \(\sqrt{9+16}=5\).

What’s the modulus of a complex number?

The modulus \(|z|\) is the distance from \(z=a+bi\) to the origin: \(|z|=\sqrt{a^2+b^2}\). For \(z=3+4i\), \(|z|=\sqrt{9+16}=5\). Modulus is always non-negative and tells you how far \(z\) is from 0 in the complex plane.

What’s the midpoint of two complex numbers?

The midpoint of \(z_1\) and \(z_2\) is \(\frac{z_1+z_2}{2}\). Add the two numbers (real parts together, imaginary parts together), then divide by 2. Example: midpoint of \(4+6i\) and \(2+2i\) is \(\frac{6+8i}{2}=3+4i\).

Why does this work the same as coordinate geometry?

Because the complex plane IS the coordinate plane with relabeled axes. Real axis = \(x\)-axis; imaginary axis = \(y\)-axis. Distance and midpoint formulas from geometry transfer directly. You can think of every complex-number geometry problem as a regular geometry problem in disguise.

Walk through a full example?

Find the distance and midpoint between \(z_1=5+3i\) and \(z_2=1-i\). Distance: \(\sqrt{(5-1)^2+(3-(-1))^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\). Midpoint: \(\frac{(5+1)+(3+(-1))i}{2}=\frac{6+2i}{2}=3+i\). Both formulas work on the points \((5,3)\) and \((1,-1)\) in the complex plane – identical math, just relabeled coordinates. Sketching helps you check: the midpoint should sit halfway along the segment.

What if one number is purely real or purely imaginary?

Treat the missing part as 0. For \(z_1=5\) (no imaginary part), \(a_1=5\) and \(b_1=0\). For \(z_2=3i\), \(a_2=0\) and \(b_2=3\). Distance and midpoint formulas still apply: distance from 5 to \(3i\) is \(\sqrt{25+9}=\sqrt{34}\).

How does this connect to polar form?

The modulus \(|z|\) is the polar \(r\). The angle from the positive real axis to the line from 0 to \(z\) is the argument \(\theta\). So \(z=r(\cos\theta+i\sin\theta)\) is the polar form, with \(r=|z|\). Distance in the complex plane connects nicely to polar geometry.

Where do these formulas show up on tests?

Pre-calc and Algebra II tests, the SAT Math 2 Subject Test, AP Pre-Calculus, AP Calculus BC, and CLEP exams. Typical problems: find the distance between two complex numbers, find a midpoint, find the locus of complex numbers a fixed distance from a given point (which traces out a circle).

Does the order of subtraction matter when finding the distance?

No. |z₁ − z₂| and |z₂ − z₁| are always equal, because the differences only change sign and squaring removes the sign. For 3 + 2i and 8 + 5i, you get 5 + 3i one way and −5 − 3i the other way. Both have modulus √34.

Why don’t you include i when you square the imaginary difference?

The distance formula treats each complex number as a point (a, b) on the plane. The imaginary part b is just a coordinate, like y in the regular distance formula. Squaring 3i would give −9 and make the distance too small. Always square the real number 3 to get 9.

How do you find the other endpoint if you know the midpoint?

Use z₂ = 2M − z₁. Double the midpoint, then subtract the endpoint you know. If the midpoint is 1 + 4i and one endpoint is 3 + 2i, then z₂ = (2 + 8i) − (3 + 2i) = −1 + 6i. Check by averaging the two endpoints.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• How to add and subtract complex numbers (resource currently unavailable)
• How to multiply and divide complex numbers (resource currently unavailable)
• How to use the Pythagorean theorem
• How to use the distance formula
• Axis of symmetry of quadratic functions