Hello everyone,
Need a bit of help.
(-2-2i)^5
I want to turn this into Cartesian form using the modulus.
Into the form
z^n=r^n(cos n0 +i sin n0)
So far I have r=2sqrt(2)
The angle ive worked out as pi/4 or 45 degrees (im not 100% confident here).
Now im a bit unsure how I put it together.
Thanks for any help!
yes r^2 = 4 + 4 = 8
so
r = sqrt 8 = 2 sqrt 2
- 2 - 2 i is in quadrant 3, 45 degrees below -x axis
Theta = pi + pi/4 = 5 pi/4
or 180 + 45 = 225 degress
so
x = 8^.5 e^(5 pi i/4)
now raise to power five
raise (8^.5)^5 = 8^2.5
multiply angle by 5
25 pi/4 = 24 pi/4 + pi/4
which is
6 pi + pi/4
which is just pi/4 because every 2 pi is all the way around
so
8^2.5 (cos 45 + i sin 45
8^2.5 (sqrt2 /2 + i sqrt 2/2)
2^7.5 (2^.5 + i 2^.5)/2
(2^8 + i 2^8)/2
2^7 + i 2^7
or
128 + 128 i
Well, that was entertaining :)
Thanks Damon, really helped.
To express the complex number (-2-2i)^5 in Cartesian form using the modulus, you need to find the modulus (r) and the angle (θ) of the complex number.
To find the modulus (r), you can use the formula:
r = |z| = sqrt(Re(z)^2 + Im(z)^2)
In this case, Re(z) is -2 and Im(z) is -2. So the modulus is:
r = sqrt((-2)^2 + (-2)^2) = sqrt(8) = 2sqrt(2)
Now, to find the angle (θ), you can use the formula:
θ = arg(z) = arctan(Im(z)/Re(z))
In this case, Im(z) is -2 and Re(z) is -2. So the angle is:
θ = arctan((-2)/(-2)) = arctan(1) = π/4 or 45 degrees
Now that you have the modulus (r) as 2sqrt(2) and the angle (θ) as π/4 or 45 degrees, you can express the complex number (-2-2i)^5 in Cartesian form using the modulus:
z^n = r^n (cos nθ + i sin nθ)
In this case, n is 5. So you have:
(-2-2i)^5 = (2sqrt(2))^5 (cos 5(π/4) + j sin 5(π/4))
Simplifying this further, you have:
(-2-2i)^5 = 32 (cos 5(π/4) + j sin 5(π/4))
Therefore, the complex number (-2-2i)^5 in Cartesian form using the modulus is:
32 (cos 5(π/4) + j sin 5(π/4))