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The Complex Plane;DeMoivre's Theorem-
converting to trigonometric form
The magnitude or modulus of z is the same as r.
We can take complex numbers given asand convert them to polar form. Recall the conversions:
RealAxis
Imaginary Axis
yixz
cosrx sinry z= rcos( )+ rsin( )i
Do Now:
Plot the complex number:
iz 3
sincos ir factor r out
RealAxis
Imaginary Axis
Remember a complex number has a real part and an imaginary part. These are used to plot complex numbers on a complex plane.
yixz
The magnitude or modulus of z denoted by z is the distance from the origin to the point (x, y).
y
x
z
22 yxz
x
y1tan
The angle formed from the real axis and a line from the origin to (x, y) is called the argument of z, with requirement that 0 < 2.
modified for quadrant and so that it is between 0 and 2
yixz
The magnitude or modulus of z is the same as r.
We can take complex numbers given asand convert them to polar form. Recall the conversions:
RealAxis
Imaginary Axis
yixz
y
x
z = r
cosrx sinry z= rcos( )+ rsin( )i
Plot the complex number: iz 3
Find the polar form of this number.
1
3
r = − 3( )2+ 1( )
2 = 4 =2
sincos ir factor r out
Shorthand:
The angle is between - and
RealAxis
Imaginary Axis
y
x
z = r
1
3
3
1tan 1 but in Quad II
6
5
6
5sin
6
5cos2
iz
=5π
6⇒ principal α =
5π
6
or "2cis56"
It is easy to convert from polar to rectangular form because you just work the trig functions and distribute the r through.
i 3
6
5sin
6
5cos2
iz
i
2
1
2
32
2
3 2
1
If asked to plot the point and it is in polar form, you would plot the angle and radius.
2
6
5 Notice that is the same as plotting
3 i 31
Convert the following:
1) 3−3i
2) −4i
3) 4(cos116
+ isin116
)
Do Now: Convert the following to trigonometric form:
z1 =32
−12
i
z2 =−1+ i
use sum formula for sinuse sum formula for cos
Replace i 2 with -1 and group real terms and then imaginary terms
irr 2121212121 sincoscossinsinsincoscos
Must FOIL these
221121 sincossincos iirr
21 2 1 2 2 1 1 2 1 2cos cos sin cos sin cos sin sinr r i i i
Let's try multiplying two complex numbers in polar form together.
1111 sincos irz 2222 sincos irz
1 2 1 1 1 2 2 2cos sin cos sinz z r i r i
212121 sincos irr
Look at where we started and where we ended up and see if you can make a statement as to what happens to the r 's and the 's when you multiply two complex numbers.
Multiply the r values and Add the Angles
Then numbers.complex twobe
sincos and sincosLet 22221111 irzirz
zz
rr
i1
2
1
21 2 1 2 cos sin
then,0 If 2 z
(This says to multiply two complex numbers in polar form, multiply the r values and add the angles)
(This says to divide two complex numbers in polar form, divide the r values and subtract the angles)
21212121 sincos irrzz
Then numbers.complex twobe
sincos and sincosLet 22221111 irzirz
zz
rr
i1
2
1
21 2 1 2 cos sin
then,0 If 2 z
21212121 sincos irrzz
212121 cisrrzz
212
1
2
1 cisr
r
z
z
If z=4 cos40o+ isin40o( ) and w=6 cos120o+ isin120o
( ),
find: (a) zw (b) z w
wzzw
iwiz
(b) (a) :find
,120sin120cos6 and 40sin40cos4 If
4 cos 40 sin 40 6 cos120 sin120zw i i
12040sin12040cos64 i
24 cos160 sin160i
multiply the r values add the angles(the i sine term will have same angle)
If you want the answer in complex coordinates simply compute the trig functions and multiply the 24 through.
wzzw
iwiz
(b) (a) :find
,120sin120cos6 and 40sin40cos4 If
4 cos 40 sin 40 6 cos120 sin120zw i i
12040sin12040cos64 i
24 cos160 sin160i
multiply the r values add the angles(the i sine term will have same angle)
If you want the answer in complex coordinates simply compute the trig functions and multiply the 24 through.
i34202.093969.024
i21.855.22
zw
i
i
4 40 40
6 120 120
cos sin
cos sin
46
40 120 40 120cos sin i
23
80 80cos sin i
divide the r values subtract the angles
In polar form we want an angle between 0° and 180° “PRINCIPAL ARGUMENT”
=
2
3cos 100o
( ) + i sin 100o( )⎡⎣ ⎤⎦
zw
i
i
4 40 40
6 120 120
cos sin
cos sin
46
40 120 40 120cos sin i
23
80 80cos sin i
divide the r values subtract the angles
In polar form we want an angle between 0° and 180° PRINCIPAL ARGUMENT
In rectangular coordinates:
ii 66.012.09848.01736.03
2
− 3 + i( )2
Today’s theorem is used to raise complex numbers to powers.
Do Now: simplify (foil)
− 3 + i( )2
= 2 − 2i 3
Today’s theorem is used to raise complex numbers to powers.
Do Now: simplify
Abraham de Moivre(1667 - 1754)
DeMoivre’s Theorem
If is a complex number,
then
z r i cos sin
z r n i nn n cos sin integer. positive a is 1 where n
You can repeat this process raising complex numbers to powers. Abraham DeMoivre did this and proved the following theorem:
z r n i nn n cos sin
This says to raise a complex number to a power, raise the r value to that power and multiply the angle by that power.
− 3 + i( )2
This theorem is used to raise complex numbers to powers.
Instead let's convert to polar form and use DeMoivre's Theorem:
CONVERT:
APPLY DEMOIVRE’S THEOREM:
CONVERT BACK:
r 3 1 4 22 2
3
1tan 1 =150
z=2cis150
z2 =22cis(2 • 150) =4cis300
4(cos 300+ isin300)=2−2i 3
− 3 + i( )4This theorem is used to raise complex numbers
to powers. It would be a lot of work to find
iiii 3333 you would need to FOIL and multiply all of these together and simplify powers of i --- UGH!Instead let's convert to polar form
and use DeMoivre's Theorem.
− 3 + i( )4
r 3 1 4 22 2
Z 4 = 2 cos56
+ isin56
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
4
This theorem is used to raise complex numbers to powers. It would be a lot of work to find
iiii 3333 you would need to FOIL and multiply all of these together and simplify powers of i --- UGH!Instead let's convert to polar form
and use DeMoivre's Theorem.
3
1tan 1 but in Quad II
=150o =
5π
6z=2cis
56
6
54sin
6
54cos24
i
− 3 + i( )4
Z 4 = 2 cos56
+ isin56
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
4
This theorem is used to raise complex numbers to powers. It would be a lot of work to find
=16 cos20π
6
⎛
⎝⎜
⎞
⎠⎟+ i sin
20π
6
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
i
2
3
2
116
=−8 − 8 3i
Example:Remember:
ConvertApply Demoivre
Convert back
1
2−i
32
⎛
⎝⎜
⎞
⎠⎟
3
1st : (Put into calculator and see what the value is)
Example:
Convert:
1
2−i
32
⎛
⎝⎜
⎞
⎠⎟
3
⇒ (1cis(−60))3
Example:
Convert:
Apply:
1
2−i
32
⎛
⎝⎜
⎞
⎠⎟
3
⇒ (1cis(−60))3
13cis(3• −60)=1cis(−180)
Convert back:
1
2−i
32
⎛
⎝⎜
⎞
⎠⎟
3
⇒
1cis(−180)=1cis(180)1(cos(180)+ isin(180)) =−1+0i =−1
Solve the following over the set of complex numbers:
13 z We get 1 but we know that there are 3 roots. So we want the complex cube roots of 1.
Using DeMoivre's Theorem , we can develop a method for finding complex roots. This leads to the following formula:
zk = rn cosn+360k
n
⎛
⎝⎜
⎞
⎠⎟+ isin
n+360k
n
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
1 , ,2 ,1 ,0 where nk
N is the root and k goes from 0 to 1 less than the root.
We find the number of roots necessary
zk = rn cosn+360k
n
⎛
⎝⎜
⎞
⎠⎟+ isin
n+360k
n
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
1 , ,2 ,1 ,0 where nk
101 22 r
zk = rn cosn+360k
n
⎛
⎝⎜
⎞
⎠⎟+ isin
n+360k
n
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
Let's try this on our problem. We want the cube roots of 1.
We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 + 0i)
tan−1 0
1
⎛
⎝⎜⎞
⎠⎟= 0 ⇒
We want cube root so use 3 numbers here
Once we build the formula, we use it first with k = 0 and get one root, then with k = 1 to get the second root and finally with k = 2 for last root.
zk = 13 cos03+360k3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360k3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥, for k=0, 1, 2
1cis0
zk =1 cos 0( )+ isin 0( )⎡⎣ ⎤⎦
z1 = 13 cos03+360 1( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 1( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
z2 = 13 cos03+360 2( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 2( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
z0 = 13 cos03+360 0( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 0( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
zk = 13 cos03+360k3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360k3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥, for k=0, 1, 2
zk =1 cos 0( )+ isin 0( )⎡⎣ ⎤⎦
z1 = 13 cos03+360 1( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 1( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
z2 = 13 cos03+360 2( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 2( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
z0 = 13 cos03+360 0( )
3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360 0( )
3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
zk = 13 cos03+360k3
⎛
⎝⎜
⎞
⎠⎟+ isin
03+360k3
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥, for k=0, 1, 2
10sin0cos1 i
=1 cos 120( ) + i sin 120( )⎡⎣ ⎤⎦= −1
2+
3
2i
=1 cos 240( ) + i sin 240( )⎡⎣ ⎤⎦= −1
2−
3
2i
ii2
3
2
1,
2
3
2
1,1 We found the cube roots of 1 were:
Let's plot these on the complex plane about 0.9
Notice each of the complex roots has the
same magnitude (1). Also the
three points are evenly spaced
on a circle. This will always be
true of complex roots.
each line is 1/2 unit
1−i 3( )
Find the cube roots of:
Find z in trig form:
1−i 3( )13
z=2cis(−60)
zk =213cis
−603
⎛
⎝⎜
⎞
⎠⎟= 23 cis(−20)
Find the cube roots of:
1−i 3( )13
z0 = 23 cis(−20)
z1 = 23 cis −20+3603
⎛
⎝⎜
⎞
⎠⎟= 23 cis100
z2 = 23 cis −20+7203
⎛
⎝⎜
⎞
⎠⎟= 23 cis220
the cube roots in trig form:
Root 1
Root 2
Root 3
1−i 3( )13
z0 = 23 cis(−20) ≈1.18−.43i
z1 = 23 cis −20+3603
⎛
⎝⎜
⎞
⎠⎟= 23 cis100 ≈−.22 +1.24i
z2 = 23 cis −20+7203
⎛
⎝⎜
⎞
⎠⎟= 23 cis220 ≈−.97−.81i
the cube roots in complex form:
Find all real and complex cube roots of -8
Z= -8 + 0iFind trig form first:
Find all real and complex roots of -8
Z= -8 + 0iFind trig form first:
Now find the 3 roots in trig form
r =8 =180z=8cis180
zk = 83 cis1803
⎛
⎝⎜
⎞
⎠⎟=2cis60
Find all real and complex roots of -8
Z= -8 + 0i
the 3 roots in trigonometric form:
zk = 83 cis1803
⎛
⎝⎜
⎞
⎠⎟=2cis60
z0 =2cis60
z1 =2cis 60 +3603
⎛
⎝⎜
⎞
⎠⎟=2cis180
z2 =2cis 60 +7203
⎛
⎝⎜
⎞
⎠⎟=2cis300
Find all real and complex roots of -8
Z= -8 + 0i
the 3 roots in complex form:
zk = 83 cis1803
⎛
⎝⎜
⎞
⎠⎟=2cis60
z0 =2cis60 =1+ i 3
z1 =2cis 60 +3603
⎛
⎝⎜
⎞
⎠⎟=2cis180 =−2
z3 =2cis 60 +7203
⎛
⎝⎜
⎞
⎠⎟=2cis300 =1−i 3
Do Now:Find all real roots of
164
Find all real and complex fourth roots of 16
Z= 16+ 0iFind trig form first:
16cis0
Write it 4 times:
Find all real and complex fourth roots of 16
Z= 16+ 0i
z1 =2cis(0 +3604
) =2cis90
z2 =2cis(0 +7204
) =2cis180
z3 =2cis(0 +10804
) =2cis270
z0 =2cis0
Find all real and complex fourth roots of 16
Z= 16+ 0i
2cis(0+3604
) =2cis90 =2i
2cis(0 +7204
) =2cis180 =−2
2cis(0 +10804
) =2cis270 =−2i
2cis0 =2
Find all real and complex 6th roots of
Z= 1- i
Find all real and complex 6th roots of
Z= 1- i
zk =(212 )
16 cis
−456
= 212 cis(−7.5)
z= 2cis(−45)
Find all real and complex 6th roots of 1- i
z0 = 212 cis(−7.5)
z1 = 212 cis −7.5 +3606
⎛
⎝⎜
⎞
⎠⎟= 212 cis52.5 ≈.64 +.84i
z2 = 212 cis −7.5 +7206
⎛
⎝⎜
⎞
⎠⎟= 212 cis112.5
z3 = 212 cis −7.5 +10806
⎛
⎝⎜
⎞
⎠⎟= 212 cis172.5
z4 = 212 cis −7.5 +14406
⎛
⎝⎜
⎞
⎠⎟= 212 cis232.5
z5 = 212 cis −7.5 +18006
⎛
⎝⎜
⎞
⎠⎟= 212 cis292.5
