[tex]z = 4-i[/tex] has a distance of [tex]\sqrt{17}[/tex].
Note - This question is about the complex numbers, in which we need to determine the norms of a given set of complex numbers. Complex numbers are numbers with the following number:
[tex]z = a+i\,b[/tex] (1)
Whose norm is defined below:
[tex]\|z\| = \sqrt{a^{2}+b^{2}}[/tex] (2)
Now we proceed to check each complex number:
i) [tex]z = 2 + i\,15[/tex]
[tex]\|z\| = \sqrt{2^{2}+15^{2}}[/tex]
[tex]\|z\| \approx 15.132[/tex]
ii) [tex]z = i\,17[/tex]
[tex]\|z\| = 17[/tex]
iii) [tex]z = 20-i\,3[/tex]
[tex]\|z\| = \sqrt{20^{2}+(-3)^{2}}[/tex]
[tex]\|z\| = \sqrt{409}[/tex]
[tex]\|z\| \approx 20.224[/tex]
iv) [tex]z = 4 - i[/tex]
[tex]\|z\| = \sqrt{4^{2}+(-1)^{2}}[/tex]
[tex]\|z\| = \sqrt{17}[/tex]
[tex]\|z\|\approx 4.123[/tex]
Therefore, [tex]z = 4-i[/tex] has a distance of [tex]\sqrt{17}[/tex].
We kindly invite to see this question on complex numbers: https://brainly.com/question/18392150