4. Find the indicated roots of these complex numbers.
a. 3√1 = 1?
b. sqrt{- 3 - 4i}
c. 4sqrt{81cis60°}
d. 5sqrt{32cis (pi/6)}
I will do two of them, you try the rest
Using De Moivre's Theorem
b.
let z = -3 - 4i
magnitude = √((-3)^2 + (-4)^2) = 5
tanØ = 180 + 53.13 or 233.13° , since -3-4i is in the third quadrant of the
Argand plane
so we can say: z = 5cis233.13°
then z^(1/2) = (-3-4i)^(1/2) = √5cis 233.13/2° = √5 cis 116.565°
There will be 2 roots, each 360/2 or 180° apart, so the roots are
√5cis 116.565° and √5cis296.565
or -1 + 2i and 1 - 2i
let's check the last one algebraically:
(1-2i)^2 = 1 - 4i + 4i^2
= 1 - 4i - 4
= -3 - 4i , just for fun check the other one .
c.
I will assume you want the fourth root of 81cis60°
let z = 81cis60°
then z^(1/4) = 81^4 cis (60/4)°
= 3cis15°
But there will be 4 roots, each 360/4° or 90° apart from each other
so add multiples of 90 to the above answer to get:
3cis15° , 3cis105°, 3cis195, and 3cis285°
(note that adding another 90° would bring us back to 3cis15°)
I don't understand your notation in a.
a. To find the indicated root of a complex number, we need to find a complex number that, when raised to the indicated power, gives us the original complex number. In this case, we are looking for a complex number that, when raised to the power of 3, gives us 1.
To find the complex number that satisfies this condition, we can express 1 in polar form. We know that the polar representation of a complex number is given by r * cis(theta), where r is the magnitude of the complex number and theta is the angle it makes with the positive real axis.
Since 1 can be written as 1 * cis(0°), we can use the formula for finding the nth root of a complex number in polar form. The formula is given by:
z^(1/n) = (r^(1/n)) * cis((theta + 2kπ)/n),
where z is the complex number, n is the root we want to find, r is the magnitude of z, theta is the argument of z, and k is an integer from 0 to n-1.
For the given problem, we have n = 3 and z = 1 * cis(0°). We plug these values into the formula:
(1 * cis(0°))^(1/3) = (1^(1/3)) * cis((0° + 2kπ)/3),
where k = 0, 1, 2.
Simplifying, we get:
1^(1/3) = 1,
so the three solutions are:
1 * cis(0°/3) = 1 * cis(0°) = 1,
1 * cis(2π/3) = -1/2 + sqrt(3)/2 * i,
1 * cis(4π/3) = -1/2 - sqrt(3)/2 * i.
Therefore, the indicated roots of the complex number 1 are 1, -1/2 + sqrt(3)/2 * i, and -1/2 - sqrt(3)/2 * i.
b. To find the square root of a complex number, we can again express it in polar form. The square root of a complex number z = r * cis(theta) is given by:
sqrt(z) = sqrt(r) * cis(theta/2) or -sqrt(r) * cis(theta/2),
where r is the magnitude of z and theta is the argument of z.
For the given problem, we have z = -3 - 4i. Let's find its magnitude and argument.
Magnitude (r):
|z| = sqrt((-3)^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5.
Argument (theta):
tan(theta) = -4/-3 = 4/3,
theta = atan(4/3) = 53.13°.
The polar form of z is therefore 5 * cis(53.13°). Now we can find the square root:
sqrt(-3 - 4i) = sqrt(5) * cis(53.13°/2) or -sqrt(5) * cis(53.13°/2).
Calculating, we get:
sqrt(5) * cis(53.13°/2) ≈ 1.58 * cis(26.57°),
-sqrt(5) * cis(53.13°/2) ≈ -1.58 * cis(26.57°),
Therefore, the indicated square roots of the complex number -3 - 4i are approximately 1.58 * cis(26.57°) and -1.58 * cis(26.57°).
c. For this problem, we have 4√(81cis60°). Let's express the complex number in rectangular form:
81 * cis(60°) = 81(cos(60°) + isin(60°)) = 81(1/2 + sqrt(3)/2 * i).
Now we can take the fourth root of this complex number:
(81 * cis(60°))^(1/4) = (81^(1/4)) * cis((60° + 2kπ)/4),
where k = 0, 1, 2, 3.
Simplifying, we get:
81^(1/4) = 3,
so the four solutions are:
3 * cis((60° + 2kπ)/4).
The four indicated roots of the complex number 4sqrt(81cis60°) are:
3 * cis(15°),
3 * cis(75°),
3 * cis(135°),
3 * cis(195°).
d. For this problem, we have 5√(32cis(π/6)). Let's express the complex number in rectangular form:
32 * cis(π/6) = 32(cos(π/6) + isin(π/6)) = 32(√3/2 + 1/2 * i).
Now we can take the fifth root of this complex number:
(32 * cis(π/6))^(1/5) = (32^(1/5)) * cis((π/6 + 2kπ)/5),
where k = 0, 1, 2, 3, 4.
Simplifying, we get:
32^(1/5) = 2,
so the five solutions are:
2 * cis((π/6 + 2kπ)/5).
The five indicated roots of the complex number 5sqrt(32cis(π/6)) are:
2 * cis(π/30),
2 * cis(π/6),
2 * cis(7π/30),
2 * cis(2π/3),
2 * cis(13π/30).