New Zealand
Level 8 - NCEA Level 3

# Distance between and Midpoints of

Lesson

### Distance between two complex numbers

While it makes no sense to think about ordering the complex numbers, it is meaningful to talk about distances between them.

The distance between the two complex numbers $z=a+bi$z=a+bi and $w=c+di$w=c+di is the non-negative real number $\left|z-w\right|=\sqrt{\left(a-c\right)^2+\left(b-d\right)^2}$|zw|=(ac)2+(bd)2 in accordance with Pythagoras' Theorem

So, for example the distance between $z=2-5i$z=25i and $w=-1-i$w=1i is given by $\left|z-w\right|=\sqrt{\left(2+1\right)^2+\left(-5+1\right)^2}=\sqrt{9+16}=5$|zw|=(2+1)2+(5+1)2=9+16=5.

### The mid-point between two complex numbers

Similarly, we can determine the midpoint of any two complex numbers $z=a+bi$z=a+bi and $w=c+di$w=c+di by using the midpoint rule from coordinate geometry.

Thus midpoint $m$m is the complex number given by $m=\frac{a+c}{2}+\frac{b+d}{2}i$m=a+c2+b+d2i.

Using the previous complex numbers, we have that the midpoint between  $z=2-5i$z=25i and $w=-1-i$w=1i is the complex number $m=\frac{2-1}{2}+\frac{-5-1}{2}i=\frac{1}{2}-3i$m=212+512i=123i

#### An example

Find the distance between the midpoint of the two complex numbers $z_1=5+12i$z1=5+12i and $z_2=7-24i$z2=724i and the complex number $z_3=14+9i$z3=14+9i.

Then the midpoint, say $z_m$zm, is the complex number $z_m=\frac{5+7}{2}+\frac{12-24}{2}i=6-6i$zm=5+72+12242i=66i. The distance between $z_m$zm and $z_3$z3 is then $\left|z_3-z_m\right|=\sqrt{\left(14-6\right)^2+\left(9+6\right)^2}=\sqrt{64+225}=17$|z3zm|=(146)2+(9+6)2=64+225=17

#### Worked Examples

##### Question 1

Find the distance between $9+4i$9+4i and $1-2i$12i.

##### Question 2

Find the midpoint between $-5+8i$5+8i and $7-2i$72i.

### Outcomes

#### M8-9

Manipulate complex numbers and present them graphically

#### 91577

Apply the algebra of complex numbers in solving problems