Find the two square roots of 5-12i in the form a+bi, where a and b are real. Mark on an Argand diagram the points P and Q representing the square roots. Find the complex number of R and S such that PQR and PQS are equilateral triangles. How do you do the last part finding R and S?

Answer 1

https://www.jiskha.com/questions/1818615/find-the-two-square-roots-of-5-2i-in-the-form-a-bi-where-a-and-b-are-real-mark-on-an

Answer 2

We could use D' Moivre's Theorem, but in this case the following method seems easier
If a+bi is the square root of 5 - 12i
then (a+bi)^2 = 5 - 12i
a^2 + 2abi + b^2 i^2 = 5 - 12i
a^2 - b^2 + 2abi = 5 - 12i
---> a^2 - b^2 = 5 and 2ab = -12 or b = -6/a
subbing
a^2 - 36/a^2 = 5
a^4 - 5a^2 - 36 = 0
(a^2 - 9)(a^2 + 4) = 0
a^2 = 9 or a^2 = -4, but in the form a+bi, both a and b are real
so a = ± 3
if a=3, b = -2
if a= -3, b = 2

√(5 - 12i) = 3-2i or -3+2i

On the Argand plane, mark 3-2i as P(3,-2) and -3+2i as Q(-3,2)
We can now treat the Argand plane just as if you had the standard x-y plane
PRQS will be a parallelogram and angles R and S are both 60°
if O is the "origin" in the Argand plane, the ROP will be a 30-60-90° right-angled triangle
slope OP = -2/3
slope of OR = 3/2
We know OP = √13, and RP = 2√13, so OR = √3√13 by comparison of ratios with the 30-60-90
from the slope of 3/2, we know tanØ = 3/2
Ø = appr 56.3099... (I stored this in my calculator)

OR = √3√13(cos 56.3099.. + i sin 56.3099..)
= 2√3 + 3√3 i ------> R = 2√3 + 3√3 i

similarly using symmetry, S = -2√3 - 3√3 i