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4 Unit Maths – Complex Numbers

Complex Laws. Let z1 = a + ib, and z2 = c + id, where a, b, c and d are real numbers.

z1+ z2=a+ib+(c+id )z1−z2=a+ ib−(c+id)z1 z2=(a+ib)(c+id)

z1z2

=a+ibc+id

Square root of a complex number.

z=√a+iblet z2=a+ib∴ x2− y2=a

xy=b2

Solve for x and y by inspection. If unable to do through inspection use the identity;

(x2+ y2 )2=( x2− y2 )2+4 x2 y2

And then perform simultaneous equations.

Conjugate. If z=x+iy then z=x−iy .

( z )=zz1+ z2=z1+z2z1−z2=z1−z2

z1 z2=z1 z2

( z1z2 )=

z1z2

Adding vectors. Complete Parallelogram Head to Tail

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V 1

V 2

V 1+V 2

V 1

V 1+V 2

V 2

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4 Unit Maths – Complex NumbersSubtracting vectors. Complete Parallelogram

Modulus-Argument form. z=x+iy=r (cosθ+isinθ)

Where;r=√ x2+ y2∧tanθ= yx

If equation is not in correct cosθ+isinθ, (eg, cosθ – isinθ) use the unit circle – just think of when the two conditions are met.

Rules of Modulus. The modulus of a complex number is its length.

If z1=z2 , thenr 1=r2|z1 z2|=|z1||z2|

|z1z2|=|z1|

|z2||zn|=|z|n

|1z|= 1|z|

|z|=|z|z z=|z|2

Rules of Argument. The argument of a complex number is the angle made with respect to the positive x-axis.

If z1=z2 , thenθ1=θ2+k 2πarg ( z1 z2 )=arg z1+arg z2

arg( z1z2 )=arg z1−arg z2

arg zn=nargz

arg( 1z )=−argz

arg z=−argz

Further vector properties. If tail is at the origin, only one letter is used. A⃗.

However, if tail is not at origin, two letters are used. A⃗B=B⃗−A⃗ .

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V 1+(−V 2 )¿V 1−V 2

V 2

−V 2

V 1V 1

V 2

V 1−V 2

V 2−V 1

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4 Unit Maths – Complex NumbersNote: goes from A to B. The arrow ‘starts’ at A and ‘ends’ at B.

Angel between two vectors.Arg( A⃗B

BD )=Arg A⃗C−Arg B⃗D (head−tail )

Rules Separate the argument like above. Draw vectors z1 and z2. Look at the heads of the vectors. Determine the direction of angle.

Note: angle is between the two heads, and the head of the angle is on the ‘head’ line.

Rotations. To rotate a complex number, z, by θ ᵒ anticlockwise, multiply z by cisθ to get zcisθ.

Reducing/Enlarging. To reduce/enlarge a complex number, by x, multiply z by x to get xz.

Purely real/imaginary. Consider the complex number; z = x + iy.

Purely imaginary When x = 0.

Whenargz=π2+kπ .

Purely real When y = 0. When argz=0+kπ .

De Moivre’s Theorem. (cosθ+isinθ)n=cosnθ+isinnθ(cosθ−isinθ)n=cosnθ−isinnθ

Trigonometry Questions such as: Find cosnp in terms of powers of cos. Start off by writing: (cosp+isinp)n = ___ On the left hand side of the equation, use De Moivre’s theorem, whilst on the right hand side,

expand normally using Pascal’s triangles. Compare the real and imaginary parts, depending on question.

Questions such as: Find cos n x in terms of multiples of x.

zn+ 1zn

=2cosnθ.

zn− 1

zn=2isinnθ.

Start off by writing: (2cosx / sinx)n=¿¿ Group powers of x with their inverse, and then use the two above equations.

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A

B

CD

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4 Unit Maths – Complex NumbersNon-root of unity. In the form; zn=complex number (note :areal number is also complex ).

Here, zn=(rcisθ)n. Turn the complex number into cis form, and then equate the equations.

Root of unity. In the form; zn=±1.

zn=(cisθ)n.

The aim is to find all the possible values of θ – and consequently find all the values of z. Must always represent roots on a unity circle. The roots form a regular polygon, and are conjugate pairs.

Factorise over complex field. ( x−z ) ( x−z )=x2−2ℜ ( z ) x+|z|2

Locus |z−(a+ib )|=c Done geometrically. Circle, centre (a, b), radius c.

|z−(a+ib)z−(c+ id)|=1

Done geometrically. Perpendicular bisector of interval joining (a, b) and (c, d).

|z−(a+ib)z−(c+ id)|=k (wherek ≠1)

Done algebraically. Change z into x + iy.

arg [ z−(a+ib)]=θ Done geometrically. A ray coming from (a, b) Let equation of locus be y = mx + b. arg ¿θ (angle with respect to the positive x-axis). Open circle at (a, b). Restrictions do apply.

arg [ z−(a+ib)z−(c+id) ]=0∨π

Done geometrically. This is a part of a straight line which passes through (a, b) and (c, d). Restrictions: can’t be (a, b) or (c, d).

arg [ z−(a+ib)z−(c+id) ]=any other angle

Draw diagram remember direction of angle. Redraw diagram, where z is the perpendicular bisector of the interval made from (a, b) and (c, d)

so that it is easier to find the equation of the circle. Draw triangles to work out radius and centre. Restrictions: above or below the line passing through (a, b) and (c, d), not including (a, b) and

(c, d).

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