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?
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
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
To find the complex numbers R and S such that PQR and PQS are equilateral triangles, we need to take into account a few properties of equilateral triangles.
An equilateral triangle has three equal sides and three equal angles. Let's call the side length of triangle PQR (or PQS) "s" as a placeholder.
First, we need to find the direction and magnitude of the complex number representing the side PQ. Since P and Q are square roots of 5-12i, we can find PQ by subtracting the coordinates of Q from the coordinates of P. Let's denote the complex number representing P as z1 and the complex number representing Q as z2.
Given z1 = a + bi and z2 = c + di, where a, b, c, and d are real numbers, we have:
z1 = √(5-12i) = a + bi
z2 = √(5-12i) = c + di
The magnitude of PQ is given by the absolute value of the difference between z1 and z2:
|PQ| = |z1 - z2|
Next, we need to determine the direction of PQ. We can find this by calculating the argument of z1 - z2 using the atan2 function.
Now, let's subdivide the complex number PQ into s equal parts. Each of these subdivisions will represent a complex number that is s units away from P and in the same direction as PQ.
To find R and S, we will use the fact that R is one-third (1/3) the length of PQ away from P, and S is two-thirds (2/3) the length of PQ away from P.
1. Find the length of PQ:
|PQ| = |z1 - z2| = |(a + bi) - (c + di)| = |(a - c) + (b - d)i|
2. Calculate the argument of PQ:
θ = atan2(b - d, a - c)
3. Determine the length of PR (and PS):
PR = (1/3) * |PQ|
PS = (2/3) * |PQ|
4. Calculate the arguments of PR and PS:
θ1 = θ
θ2 = θ
5. Find R and S using the magnitude and arguments:
R = P + PR * (cos(θ1) + sin(θ1)i)
S = P + PS * (cos(θ2) + sin(θ2)i)
By substituting the values of PR, PS, θ1, and θ2 into the formulas for R and S, you can find the complex numbers R and S.
Remember to mark points P, Q, R, and S on the Argand diagram to represent the equilateral triangles PQR and PQS.
To find the complex numbers R and S such that PQR and PQS are equilateral triangles, we can use some geometric properties of equilateral triangles.
Let's first find the complex number representing P. The square root of 5-12i can be found as follows:
1. Express 5-12i in the form r(cosθ + isinθ):
5-12i = √(5^2 + (-12)^2) * [cos(arctan(-12/5)) + isin(arctan(-12/5))]
2. Simplify the expression:
5-12i = √(169) * [cos(-67.38°) + isin(-67.38°)]
3. Write the square root in the form a + bi:
5-12i = 13 * [cos(-67.38°) + isin(-67.38°)]
Here, a = 13 * cos(-67.38°) and b = 13 * sin(-67.38°).
Now, to find the complex numbers R and S that form equilateral triangles with P:
1. Rotate P 60 degrees counterclockwise around the origin to get R:
R = P * e^(i * (60°))
2. Rotate P 60 degrees clockwise around the origin to get S:
S = P * e^(-i * (60°))
In this case, e represents Euler's number, approximately equal to 2.71828.
Now, let's calculate R and S using these expressions:
R = P * e^(i * (60°))
= (a + bi) * e^(i * (60°))
= [(a * cos(60°) - b * sin(60°)) + (a * sin(60°) + b * cos(60°))] * e^(i * (60°))
Similarly,
S = P * e^(-i * (60°))
= (a + bi) * e^(-i * (60°))
= [(a * cos(60°) + b * sin(60°)) + (-a * sin(60°) + b * cos(60°))] * e^(-i * (60°))
By substituting the values of a and b into the above expressions, you can find the exact complex numbers R and S.
After finding R and S, you can plot the points P, Q, R, and S on an Argand diagram to visualize the equilateral triangles formed by PQR and PQS.