maths mindmaps
, ARGAND DIAGRAMS
3
modulus of z
S IM
2 =
1 + 2i
i=
2 = X +
i
yi
= -
1 (5 4i)(3 2i)
+ - =
15-10i
=
:
15 +
23 + 2 ;
+
2i
12i
+ 8
+ (-2x4xxF)
x
=
+
yz Re
argument of z = O = COMPLEX CONJUGATES
ππ
tan() O yi
S
= 2 =
x+
↳
when multiplied ,
I- yi a real
*
2 =
x -
get number
&
Cartesian form modulus form ①
↑
rationalising denominator
yi
z
2= x + z= r(c0SO + isinG)
X = rCOSO make one
s
rsinO simultaneous equations
~
complex
=
y
number the same
I
g
angle between
rector S in both equations &
Subtract
10 : & positive real
S
V
axis Xi x3 x (Iti)
2
A
~
radians
L
T =
1800 degreese radians
(2 22) (2 112a)
arg(2 zi) arga argee
=
,
= + . .
3600
.
2π =
arg() =
argz ,
-arge (i) = *
*
·
90
°=
=
50 radians
LOCK S
Perpendicular HALF LINES arg(2-2) =
distance between
2= x +
yi 2, =
X+ yi 12-2 ) ,
e
fixed point
>
fixed point variable point
& variable point bisector arg(x yi 2) = + -
circles 12-al-ra= centre r= radius
z = x+ yi 12-4) =
12- 2i) distancefram 24 arg(2-ao arg((x 2) + yi) = -
(2 (2 +3i)) (x + (y2)i)
same dis from ze Zi tano tan =
((x 4) yi)
↓
-
= 5 -
+ =
tan
22
3i)
2
5
(x + yi (x 4) + y2
,
Zi =
-
2 -
= 5 -
= x + (y 2)2
-
X
O
-----
y
=
53x 25
((x 2) (y 3)i) 5 4
-
-
+ - =
-
8x + 16 =
4y +4 3/2
Circle centre 1T/
2x 3 3 X2
(X 2) + (y 3)" y
-
25 (2, 3) radius 5
= -
- -
=
,
(2 0,
&
, 3x2 2x + 4 4x" 1(x 26x
g
+ 15x = 0
=2
-
- -
axi + bx + C
p q + if f(z) = 0
has roots pag
roots p q 1 + 2i factor f(z )
*
,
X = is a . then =
G
p +q = find pi + g2
pq
= find cubic a
: X = 1-2i is a factor
polynomial with roof
·
.
=
.
pq
·
(p + q) p + q2-2pq =
3 & 4+ i (X -(1 2i))(x + -
(1 -
zi)
ax + bx d
roots =
3 , 4+ , 4-i
i
+ (x +
X X(l + zi) X(l 2i) + 5
(3)2 2(5) E B 2
:
i)
- -
-
(x 3)(x (4 + i))(x (4
= -
=
roots p g r
-
-
- -
X2 - 2x + 5 -> complex quadratic factor
-
, ,
p + q+ r= (x 3)(x2 8x + 17)
find
+
-
4x2 3x quadratic factor
-
real
·
- -
pq gr + +
pr = x2 2x+ 5 4x" 1(x + 26x2- 15x
X (lx 4(x 51
&
=
+ -
-
-
-
par = ·
+= 4x" 8x3 + 20x3
-
6x
3
0 3x + 15x
G
- -
ax" + bx (x + dx 6x2 15X
=
3x*
ROOTS OF
+ + e -
+ -
4/5
i
·
roots p g , r, S
,
& g O
p + q + r+s = POLYNOMIALS
.
2
x(x2 - 2x + 5)(4x 3) 3
=
=
↓
-
pa + pr + ps + gr + qs +s
-
X(4x 3)(X (1 2i))(x (1 + (i)
=
TRANSFORMING
- -
=
- -
par +
pqs +
prs + srq EQUATIONS
parse 3x3 x-
+ 2x + 5 =
0
: roots =
1 + 2i , 1-2i ,
0 ,
34
has roots p g ,r
(a + bi)(a bi)
,
a + b2 =
-
find cubic equation with roots (p-2) (G-2) , ,
(r-2)
METHOD 2
p + q+ r = 5x2 + bx + c = 0
p 2 +
q 2 +r z p + g+ r 6 METHOD
-
=
1
- -
-
has root 4 + 7 :
=
- -
6 =
7= X p q r =
, ,
find b & C
2 r 2
u p z g
-
-
-
= , ,
4 + Ti
pa + gr + rp
=
= (p 2)(q 2) (q 2)(r z)
- -
+ -
-
+ (r z)(p
-
2)
-
u +2 =
p q , ,
w
roots =
,
4-7i
=
pq + gr + rp -
4(p + a + r) + 12 3(u + 2) -
(u + 2)2 + 2(u + 2) +5 =
0
4 + Ti +
=-
4 Ti -
8: b 40
= 4(5) 3 7
=
= -
+ 12 = =
= 3u3 + 1742 + 344 + 29 =
0
(4 + Ti)(4 Ti) -
=
par = (p-2)(q 2) (r -2) -
=
par-2(pa + ar +
rp) -
8
a 3 b =1 34d =
2 16
--
=
: -
2(z) -
8 = =
:
3x + 17x + 34x + 29 65 =
5 c 325 =
, ARGAND DIAGRAMS
3
modulus of z
S IM
2 =
1 + 2i
i=
2 = X +
i
yi
= -
1 (5 4i)(3 2i)
+ - =
15-10i
=
:
15 +
23 + 2 ;
+
2i
12i
+ 8
+ (-2x4xxF)
x
=
+
yz Re
argument of z = O = COMPLEX CONJUGATES
ππ
tan() O yi
S
= 2 =
x+
↳
when multiplied ,
I- yi a real
*
2 =
x -
get number
&
Cartesian form modulus form ①
↑
rationalising denominator
yi
z
2= x + z= r(c0SO + isinG)
X = rCOSO make one
s
rsinO simultaneous equations
~
complex
=
y
number the same
I
g
angle between
rector S in both equations &
Subtract
10 : & positive real
S
V
axis Xi x3 x (Iti)
2
A
~
radians
L
T =
1800 degreese radians
(2 22) (2 112a)
arg(2 zi) arga argee
=
,
= + . .
3600
.
2π =
arg() =
argz ,
-arge (i) = *
*
·
90
°=
=
50 radians
LOCK S
Perpendicular HALF LINES arg(2-2) =
distance between
2= x +
yi 2, =
X+ yi 12-2 ) ,
e
fixed point
>
fixed point variable point
& variable point bisector arg(x yi 2) = + -
circles 12-al-ra= centre r= radius
z = x+ yi 12-4) =
12- 2i) distancefram 24 arg(2-ao arg((x 2) + yi) = -
(2 (2 +3i)) (x + (y2)i)
same dis from ze Zi tano tan =
((x 4) yi)
↓
-
= 5 -
+ =
tan
22
3i)
2
5
(x + yi (x 4) + y2
,
Zi =
-
2 -
= 5 -
= x + (y 2)2
-
X
O
-----
y
=
53x 25
((x 2) (y 3)i) 5 4
-
-
+ - =
-
8x + 16 =
4y +4 3/2
Circle centre 1T/
2x 3 3 X2
(X 2) + (y 3)" y
-
25 (2, 3) radius 5
= -
- -
=
,
(2 0,
&
, 3x2 2x + 4 4x" 1(x 26x
g
+ 15x = 0
=2
-
- -
axi + bx + C
p q + if f(z) = 0
has roots pag
roots p q 1 + 2i factor f(z )
*
,
X = is a . then =
G
p +q = find pi + g2
pq
= find cubic a
: X = 1-2i is a factor
polynomial with roof
·
.
=
.
pq
·
(p + q) p + q2-2pq =
3 & 4+ i (X -(1 2i))(x + -
(1 -
zi)
ax + bx d
roots =
3 , 4+ , 4-i
i
+ (x +
X X(l + zi) X(l 2i) + 5
(3)2 2(5) E B 2
:
i)
- -
-
(x 3)(x (4 + i))(x (4
= -
=
roots p g r
-
-
- -
X2 - 2x + 5 -> complex quadratic factor
-
, ,
p + q+ r= (x 3)(x2 8x + 17)
find
+
-
4x2 3x quadratic factor
-
real
·
- -
pq gr + +
pr = x2 2x+ 5 4x" 1(x + 26x2- 15x
X (lx 4(x 51
&
=
+ -
-
-
-
par = ·
+= 4x" 8x3 + 20x3
-
6x
3
0 3x + 15x
G
- -
ax" + bx (x + dx 6x2 15X
=
3x*
ROOTS OF
+ + e -
+ -
4/5
i
·
roots p g , r, S
,
& g O
p + q + r+s = POLYNOMIALS
.
2
x(x2 - 2x + 5)(4x 3) 3
=
=
↓
-
pa + pr + ps + gr + qs +s
-
X(4x 3)(X (1 2i))(x (1 + (i)
=
TRANSFORMING
- -
=
- -
par +
pqs +
prs + srq EQUATIONS
parse 3x3 x-
+ 2x + 5 =
0
: roots =
1 + 2i , 1-2i ,
0 ,
34
has roots p g ,r
(a + bi)(a bi)
,
a + b2 =
-
find cubic equation with roots (p-2) (G-2) , ,
(r-2)
METHOD 2
p + q+ r = 5x2 + bx + c = 0
p 2 +
q 2 +r z p + g+ r 6 METHOD
-
=
1
- -
-
has root 4 + 7 :
=
- -
6 =
7= X p q r =
, ,
find b & C
2 r 2
u p z g
-
-
-
= , ,
4 + Ti
pa + gr + rp
=
= (p 2)(q 2) (q 2)(r z)
- -
+ -
-
+ (r z)(p
-
2)
-
u +2 =
p q , ,
w
roots =
,
4-7i
=
pq + gr + rp -
4(p + a + r) + 12 3(u + 2) -
(u + 2)2 + 2(u + 2) +5 =
0
4 + Ti +
=-
4 Ti -
8: b 40
= 4(5) 3 7
=
= -
+ 12 = =
= 3u3 + 1742 + 344 + 29 =
0
(4 + Ti)(4 Ti) -
=
par = (p-2)(q 2) (r -2) -
=
par-2(pa + ar +
rp) -
8
a 3 b =1 34d =
2 16
--
=
: -
2(z) -
8 = =
:
3x + 17x + 34x + 29 65 =
5 c 325 =
