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RSB [31]
3 years ago
9

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

Mathematics
2 answers:
Amanda [17]3 years ago
7 0

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.

guapka [62]3 years ago
6 0
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

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