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

a. 2 + 15i
b. 17 + i
c. 20 - 3i
d. 4 - i

User Bosnjak
by
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2 Answers

0 votes

Answer:

d 4-i

Explanation:

User SkaveRat
by
8.7k points
5 votes
The distance of a complex number a+bi from the origin on the complex plane is given by the formula:

√(a^2+b^2)

a.

2+15i \\ a=2 \\ b=15 \\ \Downarrow \\ d=√(2^2+15^2)=√(4+225)=√(229)

b.

17+i \\ a=17 \\ b=1 \\ \Downarrow \\ d=√(17^2+1^2)=√(289+1)=√(290)

c.

20-3i \\ a=20 \\ b=-3 \\ \Downarrow \\ d=√(20^2+(-3)^2)=√(400+9)=√(409)

d.

4-i \\ a=4 \\ b=-1 \\ \Downarrow \\ d=√(4^2+(-1)^2)=√(16+1)=√(17)

The answer is D.
User Breed Hansen
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