Find the fourth roots of the complex number in square brackets below?

Question

w = [ ( 1 + i ) / ( 2 - i ) ] ^ 0.25

If anyone can manage that I would appreciate it but its soo hard, so thanks to anyone who has a go.

Thanks guys, I really appreciate it

Answer 1

w = [ ( 1 + i ) / ( 2 - i ) ] ^ 0.25
Simplify by multiplying Num and Denom by (2+i) getting:
w = (.2 + .6i)^.25
Now convert to the tigonometric form of a complex number:
w = (cos 71.565 +i sin71.565)^.25
Now use De Moivre's theorm for fractional exponents:
(cosz +isinz)^1/n = cos[(z +2k*pi)/n]+i sin[(z +2k*pi)/n]
Plug z= 71.565 degrees and n=4 in the above equation and you will get your fourth roots.

Answer 2

First simplify the number in brackets by multiplying top and bottom by the conjugate of the bottom:

[(1+i)(2+i)/(2-i)(2+i)] = [2+3i+i²]/[4-i²] = (1+3i)/5 = 1/5 + 3i/5

so now you want the 1/4th power of that, so you use DeMoivre's Theorem. First convert 1/5 + 3i/5 to polar form. Picture a right triangle, base 1/5 and height 3/5. Then tan x = (3/5)/(1/5) = 3, so x = arctan 3 = 1.249. Also, √[(1/5)²+(3/5)²] = √(10/25) = (√10)/5.

So in polar form our number is (√10)/5 (cos arctan 3 + i sin arctan 3), and the 1/4 root of it is (√10)/20 [cos (1/4 arctan 3 + i sin (1/4 arctan 3)]. That's it exactly, and since arctan 3 is not a nice number, everything else is just calculator work. So why not skip the fuss and just ask the calculator to do the whole thing? You get 0.8466 + .2740i.

Answer 3

I'm using w to represent
w = [ ( 1 + i ) / ( 2 - i ) ] rather than the fourth root. Then we'll find w^.25.

First multiply top and bottom by 2 + i, to get
w = (1+3i)/5

Now modulus = sqrt((1/5)^2 + (3/5)^2) = sqrt(10)/5, call this r.

w = r(1/5r + (3/5r)*i)
= r(cosA + i sinA) where A = arcos (1/5r) which is arcos(1/sqrt10)

Hence w^.25 = r^.25(cos t + i sin t)
where t = A/4, A/4 + p1/2, A/4 + pi, A/4 + 3*pi/2.

The rest is calculator work.

Using Excel, the values I get for A are
1.2490457722.8198420994.3906384265.961434753
and r = 0.632455532

These give the following values for w^.25:

0.282005448 + 0.846016345*i
-0.846016345 + 0.282005448*i
-0.282005448 - 0.846016345*i
0.846016345 - 0.282005448*i

Answer 4

permit 5 + 12i = a + bi we are in a position to assert that (x + iy)² = 5 + 12i increasing the brackets on the LHS: x² + 2xyi + i²y² = 5 + 12i x² + 2xy*i - y² = 5 + 12i Equating coefficients we are in a position to coach that x² - y² = 5 and 2xy = 12, as a result xy = 6, as a result x = 6/y (6/y)² - y² = 5 36/y² - y² = 5 y^4 + 5y² - 36 = 0 (y² + 9)(y² - 4) = 0 y² = -9 or 4 y = ±2 or ±3i x = ±3 or ±2i as a result the sq. root of 5 + 12i = ±(3 + 2i)

Answer 5

This is much easier in polar notation.

First, simplify the complex fraction.
(1 + i)/(2 - i) = (sqrt2 @ 45°)/(sqrt5 @ 26.565°) = sqrt(0.2) @ 18.435° = 0.4472 @ 18.435°

The primary 4th root of this is:
0.4472^(1/4) @ 18.435/4 °
The other three roots are at 90° increments to this one.

The roots are:
0.8178 @ 4.6087°
0.8178 @ 94.6087°
0.8178 @ 184.6087°
0.8178 @ 274.6087°

I assume you can convert those back into Cartesian coordinates if you need to.

Answer 6

(1) Finding the complex conjugate of w,

w = [(1+i)/(2-i)]^0.25
= (0.2 + 0.6i)^0.25 OR (1/5 + 3i/5)^0.25

(2) Finding the modulus of w,

|w| = √of 0.2^2 + 0.6^2
= √ of 0.4

(3) Finding arg of w,

tan θ = 0.6/0,2
therefore θ = arctan 3
= 1.25 rad

(4) Use de Moivre's Theorem,
w^0.25 = (√of 0.4)^0.25 x е^i[0.25x(θ+2kπ)
= 0.8918 x е^i[0.25x(1.25+2kπ)]

when
k=0, w^0.25 = 0.8918 x е^i[0.3125] = 0.8486 + 0.2742i

k=1, w^0.25 = 0.8918 x е^i[0.25x(1.25+2π)] = -0.2742 + 0.8486i

k=2, w^0.25 = 0.8918 x е^i[0.25x(1.25+4π)] = -0.8486 - 0.2742i

k=3, w^0.25 = 0.8918 x е^i[0.25x(1.25+6π)] = 0.2742 - 0.8286

Those are your four roots

Answer 7

The approximated answer is 0.848654 + 0.273965i