Question

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

Answer 1

Answer:

25

Step-by-step explanation:

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

Answer 2

Answer:

25.

Step-by-step 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| =  $\sqrt{(2-9)^{2}+(-3-21)^{2}}$

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

=  $\sqrt{49+576}$

=  $\sqrt{625}$

= 25.