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

OptionD

Step-by-step explanation:

4-i option d is the answer

REcall that when z =a+ib for a complex number distance from origin is

modulus z

|z|=

Apply the above rule for each option

Option 1.

Distance = not our answer

Similarly for option b

Distance = \sqrt{17^2+1^2}=\sqrt{290} not our answer

Option c:

\sqrt{20^2+3^2}=\sqrt{409}not matching with given value

Option d

Distance=\sqrt{4^2+1^2}=\sqrt{17}=given answer

    Hence option d is right

the image of (6, 9) under a dilation is (4, 6). the scale factor is

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.