Question

Which complex number has a distance of StartRoot 17 EndRoot from the origin on the complex plane? 2 15i 17 i 20 – 3i 4 – i

Answer 1

$z = 4-i$ has a distance of $\sqrt{17}$.

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:

$z = a+i\,b$ (1)

Whose norm is defined below:

$\|z\| = \sqrt{a^{2}+b^{2}}$ (2)

Now we proceed to check each complex number:

i) $z = 2 + i\,15$

$\|z\| = \sqrt{2^{2}+15^{2}}$

$\|z\| \approx 15.132$

ii) $z = i\,17$

$\|z\| = 17$

iii) $z = 20-i\,3$

$\|z\| = \sqrt{20^{2}+(-3)^{2}}$

$\|z\| = \sqrt{409}$

$\|z\| \approx 20.224$

iv) $z = 4 - i$

$\|z\| = \sqrt{4^{2}+(-1)^{2}}$

$\|z\| = \sqrt{17}$

$\|z\|\approx 4.123$

Therefore, $z = 4-i$ has a distance of $\sqrt{17}$.

We kindly invite to see this question on complex numbers: brainly.com/question/18392150