This has been answered already.
I assume you can at least find P and Q.
Then you know that the angles involved are all 60 degrees, so just treat it like one of those navigation problems involving distances and bearings.
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?
4 answers
Using De Moivre's Theorem
let z = 5-12i
the r value is √(25+144) = 13
tanØ = -12/5
Ø = 292.619 ....°
so
z = 13cis292.619... (stored in calculators for accuracy), (cisØ is shortform for cosØ+i sinØ )
z^(1/2) = √13cis146.3099.. or √13cis(146.3099+180)
= -3 + 2i or 3 - 2i
√(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
let z = 5-12i
the r value is √(25+144) = 13
tanØ = -12/5
Ø = 292.619 ....°
so
z = 13cis292.619... (stored in calculators for accuracy), (cisØ is shortform for cosØ+i sinØ )
z^(1/2) = √13cis146.3099.. or √13cis(146.3099+180)
= -3 + 2i or 3 - 2i
√(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
I gave him a cut-and-paste copy of my previous solution.
Good question, like the way it comes out.
Good question, like the way it comes out.
This is the 4th time this same person has posted this — and each under a different name (as if we couldn't figure it out!). =(
Here are the other 3:
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
https://www.jiskha.com/questions/1818622/find-the-two-square-roots-of-5-12i-in-the-form-a-bi-where-a-and-b-are-real-mark-on-an
https://www.jiskha.com/questions/1818617/find-the-two-square-roots-of-5-2i-in-the-form-a-bi-where-a-and-b-are-real-mark-on-an
Here are the other 3:
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
https://www.jiskha.com/questions/1818622/find-the-two-square-roots-of-5-12i-in-the-form-a-bi-where-a-and-b-are-real-mark-on-an
https://www.jiskha.com/questions/1818617/find-the-two-square-roots-of-5-2i-in-the-form-a-bi-where-a-and-b-are-real-mark-on-an
2 answers