Complex Numbers

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Distance between and Midpoints of

Lesson

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`+`b``i` and $w=c+di$`w`=`c`+`d``i` is the non-negative real number $\left|z-w\right|=\sqrt{\left(a-c\right)^2+\left(b-d\right)^2}$|`z`−`w`|=√(`a`−`c`)2+(`b`−`d`)2 in accordance with Pythagoras' Theorem.

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

Similarly, we can determine the midpoint of any two complex numbers $z=a+bi$`z`=`a`+`b``i` and $w=c+di$`w`=`c`+`d``i` 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`+`c`2+`b`+`d`2`i`.

Using the previous complex numbers, we have that the midpoint between $z=2-5i$`z`=2−5`i` and $w=-1-i$`w`=−1−`i` is the complex number $m=\frac{2-1}{2}+\frac{-5-1}{2}i=\frac{1}{2}-3i$`m`=2−12+−5−12`i`=12−3`i`.

Find the distance between the midpoint of the two complex numbers $z_1=5+12i$`z`1=5+12`i` and $z_2=7-24i$`z`2=7−24`i` and the complex number $z_3=14+9i$`z`3=14+9`i`.

Then the midpoint, say $z_m$`z``m`, is the complex number $z_m=\frac{5+7}{2}+\frac{12-24}{2}i=6-6i$`z``m`=5+72+12−242`i`=6−6`i`. The distance between $z_m$`z``m` and $z_3$`z`3 is then $\left|z_3-z_m\right|=\sqrt{\left(14-6\right)^2+\left(9+6\right)^2}=\sqrt{64+225}=17$|`z`3−`z``m`|=√(14−6)2+(9+6)2=√64+225=17.

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

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

Manipulate complex numbers and present them graphically

Apply the algebra of complex numbers in solving problems