MTH 441
WS01: Finding Square and Cube Roots of Complex Numbers V2
Fall 2016
Let ! z = a + bi be a complex number. Then ! z = a 2 + b 2 = the length of z as a vector emanating from the origin
of the complex plane. Now,
(
(
)
)
! z = z a z + b z i = z cos(θ ) + sin (θ ) i where ! cos(θ ) = a z , sin(θ ) = b z .
The two square roots of z are
! r1 =
z ( cos(θ 2) + sin (θ 2 ) i ) , and r2 = −r1 =
z ( cos(θ 2 + π ) + sin (θ 2 + π ) i ) since
! cos (θ 2 + π ) = cos (θ 2 ) cos(π ) + sin (θ 2 ) sin(π ) = − cos (θ 2 ) and
! sin (θ 2 + π ) = cos (θ 2 ) sin(π ) + sin (θ 2 ) cos(π ) = − sin (θ 2 ) .
This means that the square roots of a complex number are separated by ! π radians or 180 degrees.
Example 1: Find the square roots of ! 4i (see the figure on the next page).
First,
⎛ 0
4 ⎞
! 4i = 4i ⎜
+
i = 4 ( 0 + 1i ) = 4 ( cos(π 2) + sin(π 2)i )
⎝ 4i 4i ⎟⎠
Thus, the square roots are
! r1 =
Check:
4 ( cos(π 4) + sin (π 4 ) i ) = 2
(
! ± 2 (1+ i )
!!
)
2
((
) (
2 2 +
))
2 2 i = 2 (1+ i ) and ! r2 = −r1 = − 2 (1+ i ) .
= 2 (1+ i ) = 2 (12 + 2i + i 2 ) = 2 (1+ 2i − 1) = 4i .
2
MTH 441
WS01: Finding Square and Cube Roots of Complex Numbers V2
Fall 2016
Next, the three cube roots of z are
! r1 =
where
z ( cos(θ 3) + sin (θ 3) i ) and ! r2 = ω r1 , r3 = ω 2 r1
3
(
! ω = −1+ −3
) 2 = ( −1+
)
(
)
ω 2 = −1− 3 i 2
3 i 2,
are the two complex cube root of 1 (see the figure on page 1). Thus, the cube roots of a complex number are
separated by ! 2π 3 radians or 120 degrees!
Example 2: Find the cube roots of ! 27i .
First,
! 27i = 27 ⋅ 0 + 27i = 27 ( cos(π 2) + sin(π 2)i )
Thus, the cube roots are
! r1 =
3
27 ( cos(π 6) + sin (π 6 ) i ) = 3
((
)
⎛ 3 + i⎞
3 2 + (1 2 ) i = 3 ⎜
and
⎝ 2 ⎟⎠
)
⎛ −1+ 3 i ⎞ ⎛ 3 + i ⎞
− 3 + 3i − i − 3 3
r2 = ω r1 = ⎜
⋅ 3⎜
= 3⋅
= − 3+i
⎟
⎟
2
4
2
⎝
⎠ ⎝ 2 ⎠
(
!
⎛ −1− 3 i ⎞ ⎛ 3 + i ⎞
− 3 − 3i − i + 3
r3 = ω 2 r = ⎜
⋅ 3⎜
= 3⋅
= −3i
⎟
⎟
2
4
⎝
⎠ ⎝ 2 ⎠
Check:
3
⎛
3 + i⎞
! r13 = ⎜ 3⋅
= 27 ⋅
2 ⎟⎠
⎝
(
3+i
8
)
3
( 3 ) + 3( 3 ) i + 3( 3 ) (i)
= 27 ⋅
3
2
8
2
+ (i)3
= 27 ⋅
)
.
3 3 + 9i − 3 3 − i
= 27i
8
MTH 441
WS01: Finding Square and Cube Roots of Complex Numbers V2
where I have used that the expansion of !
with ! u =
(
3+i
)
3
Fall 2016
is given by the formula ! ( u + v ) = u 3 + 3u 2 v + 3uv 2 + v 3
3
3 and ! v = i . Therefore,
( ) r = (ω )
! r2 3 = ω 3r13 = 1⋅ 27i = 27i, and r3 3 = ω 2
3
3 2
3
1
r13 = 12 ⋅ 27i = 27i .
!!
Example 3: Find the square roots of ! z = 3 + 4i .
First, since ! 3 + 4i = 32 + 4 2 = 25 = 5 ,
⎛3 4 ⎞
+ i = 5 ( cos(θ ) + sin(θ )i )
⎝ 5 5 ⎟⎠
! 3 + 4i = 5 ⎜
where
! θ = arctan ( 4 3) = 53.13010235415598 degrees = 0.9272952180016122 radians .
Then the square roots are
!
± 5 ( cos(0.92729521800161225 2) + sin(0.9272952180016122 2)i )
= ± 5 ( 0.8944271909999159+0.4472135954999579 i) = ±(2 + i)
Check:
! ⎡⎣ ± ( 2 + i ) ⎤⎦ = 4 + 4i + (i)2 = 3 + 4i .
2
!!
Example 4: Find the cube roots of ! z = 3 + 4i .
Using the representation of ! 3 + 4i found in Example 3, the cube roots are
r1 = 3 5 ( cos(0.92729521800161225 3) + sin(0.9272952180016122 3)i )
!
= 3 5 ( 0.9526082218220564 + 0.3041998943409083 i )
= 1.6289371459221758 + 0.5201745023045458i,
! r2 = ω r1 = –1.2649529063577516 + 1.1506136983844504 i , and
! r3 = ω 2 r1 = –0.36398423956442405 – 1.670788200688996i
Check for ! r1 :
(1.6289371459221758 + 0.5201745023045458i )3 =
= (1.6289371459221758 ) + 3⋅ (1.6289371459221758 ) ⋅ ( 0.5201745023045458i )
3
!
2
+ 3⋅ (1.6289371459221758 ) ⋅ ( 0.5201745023045458i ) + ( 0.5201745023045458i )
2
3
= ( 4.32228083183271− 1.3222808318327108 ) + ( 4.140749603778405 − 0.140749603778406 ) i
= 2.9999999999999996 + 3.9999999999999987i ≈ 3 + 4i
The verifications for the other two cube roots are left to the student!
!!
MTH 441
WS01: Finding Square and Cube Roots of Complex Numbers V2
Fall 2016
Problems: For each of the following, find the indicated roots and make an accurate graph showing the given
complex number and its roots. NOTE NEATNESS COUNTS!
1. The square roots of ! z = −64i .
2. The cube roots of ! z = −64i .
3. The square roots of ! z = 5 + 12i .
4. The cube roots of ! z = 2 + 11i .