which complex number has a distance of $sqrt{17}$ from the origin on the complex plane?\n$2 + 15i$\n$17 +…

Answer:

D. $4 - i$

Explanation:

Step1: Recall distance formula for complex numbers

The distance $d$ of a complex number $z=a + bi$ from the origin in the complex - plane is given by $d=\sqrt{a^{2}+b^{2}}$.

Step2: Check option A

For $z = 2+15i$, $d=\sqrt{2^{2}+15^{2}}=\sqrt{4 + 225}=\sqrt{229}\neq\sqrt{17}$.

Step3: Check option B

For $z = 17 + i$, $d=\sqrt{17^{2}+1^{2}}=\sqrt{289 + 1}=\sqrt{290}\neq\sqrt{17}$.

Step4: Check option C

For $z = 20-3i$, $d=\sqrt{20^{2}+(-3)^{2}}=\sqrt{400 + 9}=\sqrt{409}\neq\sqrt{17}$.

Step5: Check option D

For $z = 4 - i$, $d=\sqrt{4^{2}+(-1)^{2}}=\sqrt{16 + 1}=\sqrt{17}$.