Question

Which complex number has a distance of √17 from the origin on the complex plane?

Answer 1

Answer:

The complex number 4-i has distance  $\sqrt{17}$ from origin.

D is correct

Step-by-step explanation:

We are given the absolute value of complex plane.

If complex number is a+ib then absolute value $\sqrt{a^2+b^2}$

We have to check the absolute value of each option and check which is equal to $\sqrt{17}$

Option A:  2+15i

$d=\sqrt{2^2+15^2}=\sqrt{4+225}=\sqrt{229}\neq \sqrt{17}$

Option B:  17+i

$d=\sqrt{17^2+1^2}=\sqrt{289+1}=\sqrt{290}\neq \sqrt{17}$

Option C:  20-3i

$d=\sqrt{20^2+3^2}=\sqrt{400+9}=\sqrt{409}\neq \sqrt{17}$

Option D:  4-i

$d=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}= \sqrt{17}$

Hence, The complex number 4-i has distance  $\sqrt{17}$ from origin.

Answer 2

Let the complex number be x + iy

Then by the pythagoras therem

17  = x^2 + y^2

D will satisfy this equation

4^2 + (-1)^2  = 17

answer is D  4 - i