Question

Find the distance between 2-3i and 9+21i

Answer 1

25

Explanation:

`d(z_1,\ z_2)=|z_1-z_2|\\\\z=a+bi\to|z|=√(a^2+b^2)\\------------------\\\\\text{We have:}\\\\z_1=2-3i,\ z_2=9+21i\\\\\text{substitute:}\\\\|(2-3i)-(9+21i)|=|2-3i-9-21i|\\\\=|(2-9)+(-3i-21i)|=|-7-24i|\\\\=√((-7)^2+(-24)^2)=√(49+576)=√(625)=25`

Answer 2

25.

Explanation:

For two complex numbers z=(
`x_(1),y_(1)`) and w=(
`x_(2),y_(2)`) we have that the distance between z and w is equal to

|z-w| =
`\sqrt{(x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2)}`

So, in this case we have that
`(x_(1),y_(1))=(2,-3)` and
`(x_(2),y_(2))=(9,21)`, then the distance

|z-w| =
`\sqrt{(2-9)^(2)+(-3-21)^(2)}`

=
`\sqrt{(-7)^(2)+(-24)^(2)}`

=
`√(49+576)`

=
`√(625)`

= 25.