Question 462738: Determine (-2+i3)^5 in the polar form and in cartesian form
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Determine (-2+i3)^5 in the polar form
---
r = sqrt(2^2+3^2) = sqrt(13)
theta = tan^-1(-3/2) = 303.69 degrees
----------------
(-2+3i)^5 = 13^(5/2)[cos(303.69)+isin(303.69)
=====================================================
and in cartesian form
(-2+3i)^5 = x + yi
x = 13^(5/2)*cos(303.69) = 338
y = 13^(5/2)*isin(303.69) = -507i
-----------------------------------------
Cheers,
Stan H.
==========
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website! Determine (-2+i3)^5 in the polar form and in cartesian form
The other tutor got the angle in the 4th quadrant instead
of the 2nd quadrant, and also forgot to multiply the angle
by 5, which also would have been wrong since he got the
angle in the wrong quadrant.
Draw the vector x + iy = -2 + i3 which is the
vector from (0,0) to (-2,3), and let its length
(magnitude, modulus, absolute value) be r, and its
angle (argument) be q:
Then we draw a perpendicular (in green) from the
tip of the vector to the x-axis:
x + iy = eiq = r⋅cis(q) = r⋅(cosq + i⋅sinq)
We calculate r and q)
r² = x² + y²
r² = (-2)² + (3)²
r² = 4 + 9
r² = 13
r = √13
tanq = y/x
tanq = (3)/(-2) = -1.5
q = 123.69° = 2.16 radians
We use:
(x + iy)n = rnei·nq = rncis(nq) = rn[cos(nq) + i·sin(nq)]
with n = 5, q = 123.69°, r = √13
(√13)5 = 609.3381656
5q = 5(123.69°) = 618.45° which
is coterminal with 258.45°
609.3381656[cos(258.45°) + i·sin(258.45°)] = -122-i·597
Edwin
|
|
|
