Question

Which complex number has a distance of V17 from the origin on the complex plane?

Answer 1

Answer:

$4-i$

Step-by-step explanation:

Use the formula for the distance of a complex number from the origin on the complex plane.

$\sqrt{a^2+b^2}$

$a+bi$

$4-i$

$\sqrt{4^2-i^2}$

$\sqrt{4^2+(-1)^2}=\sqrt{16+1}=\sqrt{17}$

The complex number that has a distance of $\sqrt{17}$ is $4-i$.