Question

Which complex number has a distance of (square root)17 from the origin on the complex plane?

A. 2+15i
B. 17+i
C. 20-3i
D. 4-i

Answer 1

The complex number which has a distance of √17 from the origin to the complex plane is:

Option: D

`4-i`

Explanation:

We know that the distance of a complex number of the form a+ib from the origin ( (0,0) or 0+0.i ) is given by:

`√(a^2+b^2)`

Hence, we will calculate this expression in each of the options and check which is equal to √17

A)

`2+15\cdot i`

Here a=2 and b=15

Hence,

`√(a^2+b^2)=√(2^2+(15)^2)\\\\√(a^2+b^2)=√(4+225)\\\\\\√(a^2+b^2)=√(229)\\eq √(17)`

Option: A is incorrect.

B)

`17+i`

Here a=17 and b=1

Hence,

`√(a^2+b^2)=√((17)^2+1^2)\\\\√(a^2+b^2)=√(290)\\eq √(17)`

Hence, option: B is incorrect.

C)

`20-3i`

Here a=20 and b= -3

`√(a^2+b^2)=√((20)^2+(-3)^2)\\\\√(a^2+b^2)=√(409)\\eq √(17)`

Hence, option: C is incorrect.

D)

`4-i`

Here a=4 and b= -1

`√(a^2+b^2)=√((4)^2+(-1)^2)\\\\√(a^2+b^2)=√(17)`

Hence, option: D is correct.

Answer 2

we are asked in the problem to determine which among the points has a distance of sqrt 17 from the origin. We can get the distance of each through square root of (a2 + b2) from the standard from a + bi. The answer to this problem is D. 4-i