Question

Which complex number has a distance of

17 from the origin on the complex plane?
2 + 15
17 + i
20-31
04-i

Answer 1

None of these.

Please make sure you have written the choices correctly and check the question please.

The problem and choices I see:

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

Explanation:

A complex number,
`a+bi`, and the point
`0+0i` has distance:

`√((a-0)^2+(b-0)^2)` by distance formula.

Simplifying this gives us:

`√(a^2+b^2)`

So we are looking for
`a \text{ and } b` in your choices so that
`√(a^2+b^2)=17`.

Let's begin.

Choice 1:
`2+15i`

`a=2`

`b=15`

So
`√(2^2+15^2)=√(4+225)=√(229)\approx 15.1327`.

So choice A is out if I interpreted it correctly.

Choice 2:
`17+i`

`a=17`

`b=1`

So
`√(17^2+1^2)=√(289+1)=√(290)\approx 17.0294`.

So choice B is out.

Choice 3:
`20-3i`

`a=20`

`b=-3`

So
`√(20^2+(-3)^2)=√(400+9)=√(409)\approx 20.2237`

So choice C is out.

Choice 4:
`4-i`

`a=4`

`b=-1`

So
`√(4^2+(-1)^2)=√(16+1)=√(17)\approx 4.1231`

So choice D is out.

So it would be none of these have a distance of 17 from the origin.

Answer 2

4 - i is the correct option from all the others.

Explanation:

4 - i is one of the options I choose.

First let's divide this into "a" and "b"

so, a = 4

and, b = -1

This will become : √4² + (-1)² = √16 + 1 = √17

Therefore the last option is correct from all the other.