Question

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

Answer 1

Answer:

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

                     Option: D

                    $4-i$

Step-by-step 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:

$\sqrt{a^2+b^2}$

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,

$\sqrt{a^2+b^2}=\sqrt{2^2+(15)^2}\\\\\sqrt{a^2+b^2}=\sqrt{4+225}\\\\\\\sqrt{a^2+b^2}=\sqrt{229}\neq \sqrt{17}$

Option: A is incorrect.

B)

$17+i$

Here a=17 and b=1

Hence,

$\sqrt{a^2+b^2}=\sqrt{(17)^2+1^2}\\\\\sqrt{a^2+b^2}=\sqrt{290}\neq \sqrt{17}$

Hence, option: B is incorrect.

C)

$20-3i$

Here a=20 and b= -3

$\sqrt{a^2+b^2}=\sqrt{(20)^2+(-3)^2}\\\\\sqrt{a^2+b^2}=\sqrt{409}\neq \sqrt{17}$

Hence, option: C is incorrect.

D)

$4-i$

Here a=4 and b= -1

$\sqrt{a^2+b^2}=\sqrt{(4)^2+(-1)^2}\\\\\sqrt{a^2+b^2}=\sqrt{17}$

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