Unit FP1: Further Pure Mathematics AS 1

A complex number combines a real part and an imaginary part: $z=a+ib$, where $a,b\in\mathbb R$ and $i^2=-1$. Its polar form records the same number by its distance $r$ from the origin and its directed angle $\theta$ from the positive real axis.

For $z=-3+3\sqrt3,i$, $$|z|=\sqrt{9+27}=6,\qquad \arg z=\frac{2\pi}{3},$$ because the point lies in quadrant II. Hence $$z=6\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right),\qquad z^*=-3-3\sqrt3,i.$$

Add and subtract complex numbers by matching real and imaginary parts. For products, expand and replace $i^2$ by $-1$. For a quotient, multiply numerator and denominator by the conjugate of the denominator so the denominator becomes real.

The conjugate identity $(c+id)(c-id)=c^2+d^2$ explains why rationalising a quotient works. The modulus-product rule can find an unknown modulus without first expanding the product; for example, if $|z_1z_2|=15$ and $|z_1|=3$, then $|z_2|=5$.

An Argand diagram plots $z=x+iy$ as the point $(x,y)$: the horizontal axis is $\operatorname{Re}z$ and the vertical axis is $\operatorname{Im}z$. The vector from the origin has length $|z|$ and direction $\arg z$.

Multiplication by $i$ makes the geometry visible algebraically: $$i(x+iy)=-y+ix.$$ Thus $(x,y)$ maps to $(-y,x)$, a $90^\circ$ anticlockwise rotation. Multiplication by $2i$ adds enlargement by factor $2$; division by $i$ rotates $90^\circ$ clockwise.

Complex roots with real coefficients often appear as conjugate points reflected in the real axis. Distances, perimeters and areas on an Argand diagram are ordinary coordinate geometry once each complex number has been placed at its real and imaginary coordinates.

A quadratic with real coefficients can have complex solutions when its discriminant is negative. Use the ordinary quadratic formula and rewrite the square root of a negative number using $\sqrt{-k}=i\sqrt{k}$ for $k>0$.

For $z^2-6z+13=0$, the discriminant is $$(-6)^2-4(1)(13)=-16.$$ Therefore $$z=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm4i}{2}=3\pm2i.$$ Substitution or the sum and product of the two roots checks the result.

For a polynomial with real—in particular integer—coefficients, every non-real root occurs with its complex conjugate. In a cubic, a known non-real root therefore supplies a quadratic factor, leaving one real linear factor.

Suppose $$f(x)=x^3-5x^2+11x-15$$ and $1+2i$ is a root. Then $1-2i$ is also a root, giving $$[x-(1+2i)][x-(1-2i)]=x^2-2x+5.$$ Division gives $$f(x)=(x^2-2x+5)(x-3),$$ so the third root is $3$.

Form the real quadratic factor from the conjugate pair; divide the cubic by it; solve the remaining linear factor; then state all three roots. Multiplying the factors back together checks that no coefficient or sign has drifted.

A quartic with real coefficients has four roots counting multiplicity. Use every supplied root to build factors: a non-real root supplies its conjugate, while each stated real root supplies one linear factor. Divide or compare coefficients to find the remaining factor.

Let $$f(x)=x^4-7x^3+13x^2+x-20,$$ with roots $2+i$ and $-1$. Real coefficients also give $2-i$, so $$[x-(2+i)][x-(2-i)]=x^2-4x+5.$$ The root $-1$ gives $x+1$. Dividing by $(x^2-4x+5)(x+1)$ leaves $x-4$, so the four roots are $$2+i,\ 2-i,\ -1,\ 4.$$

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$ with $a\ne0$, their sum and product can be read directly from the coefficients. There is no need to solve the quadratic first.

The result comes from $$a(x-\alpha)(x-\beta)=a\bigl[x^2-(\alpha+\beta)x+\alpha\beta\bigr].$$ Comparing the coefficients of $x$ and the constant term with $ax^2+bx+c$ gives the two relationships.

For $2x^2-3x+7=0$, $$\alpha+\beta=-\frac{-3}{2}=\frac32,\qquad \alpha\beta=\frac72.$$ These values remain valid whether the roots are real or complex.

A symmetric expression is unchanged when $\alpha$ and $\beta$ are swapped. Such expressions can often be rewritten using only $S=\alpha+\beta$ and $P=\alpha\beta$, so the individual roots never need to be found.

For the roots of $2x^2-3x+7=0$, $S=3/2$ and $P=7/2$. Hence $$\alpha^2+\beta^2=\left(\frac32\right)^2-2\left(\frac72\right)=-\frac{19}{4},$$ and $$\alpha^3+\beta^3=\left(\frac32\right)^3-3\left(\frac72\right)\left(\frac32\right)=-\frac{99}{8}.$$

First replace every paired sum, product or reciprocal by $S$ and $P$; then substitute the coefficient values; finally simplify exact fractions. If a higher power appears, build it from a lower symmetric identity rather than expanding unknown roots separately.

To form a quadratic whose roots are transformed versions $u$ and $v$ of $\alpha$ and $\beta$, calculate their new sum $S'=u+v$ and product $P'=uv$. The required monic equation is then $x^2-S'x+P'=0$.

For the roots of $2x^2-3x+7=0$, $S=3/2$ and $P=7/2$. New roots $1/\alpha$ and $1/\beta$ have $$S'=\frac SP=\frac37,\qquad P'=\frac1P=\frac27.$$ Thus $$x^2-\frac37x+\frac27=0,$$ or, with integer coefficients, $$7x^2-3x+2=0.$$

Write the two new roots explicitly; derive their sum and product before inserting numbers; form $x^2-S'x+P'=0$; then multiply through by the least common denominator. A non-zero multiple represents the same quadratic equation, so simplify to integer coefficients when requested.

Numerical root methods approximate a solution of $f(x)=0$ when exact algebra is unavailable or inconvenient. First locate a root: if $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$, then at least one root lies between $a$ and $b$.

For $f(x)=x^3-x-1$, $f(1)=-1$ and $f(2)=5$, so continuity gives a root in $[1,2]$. Since $f(1.5)=0.875$, one bisection gives $[1,1.5]$; since $f(1.25)=-0.296875$, a second gives $[1.25,1.5]$. The interval width is now $0.25$.

Using $x_0=1.3$ for the same function, $f'(x)=3x^2-1$. One Newton-Raphson step gives $$x_1=1.3-\frac{1.3^3-1.3-1}{3(1.3)^2-1}=1.325307\ldots.$$ Keep unrounded values inside the calculation, then round the requested approximation only at the end.

Bisection is slower but keeps a certified bracket. Linear interpolation usually improves a bracket with one straight-line estimate. Newton-Raphson can converge quickly from a suitable starting value, but it does not preserve a bracket and a poor start or a near-zero derivative may send the iteration away from the intended root.

The FP1 standard conics are a parabola with its vertex at the origin and axis along the $x$-axis, and a rectangular hyperbola whose asymptotes are the coordinate axes. Their constants control scale, not a translation of the centre or vertex.

For $y^2=12x$, comparison with $y^2=4ax$ gives $a=3$. For $xy=25$, comparison with $xy=c^2$ gives $c=5$ or $c=-5$; the curve depends on $c^2$, so the conventional positive scale is $|c|=5$.

A parameter replaces the two coordinates of a point by one variable. Substitution verifies that the parameterised point lies on the conic and lets the same algebra describe every allowed point.

On $y^2=12x$, $a=3$. At $t=2$, the point is $(3\cdot2^2,2\cdot3\cdot2)=(12,12)$. Conversely, a point with $y=12$ has $t=y/(2a)=2$. On $xy=25$, taking $c=5$ and $t=-2$ gives $(-10,-5/2)$, whose coordinate product is $25$.

A parabola is the locus of points whose distance from a fixed point, the focus, equals their perpendicular distance from a fixed line, the directrix.

For $P=(x,y)$ on $y^2=4ax$ with $a>0$, $$PF^2=(x-a)^2+y^2=(x-a)^2+4ax=(x+a)^2.$$ Since the perpendicular distance from $P$ to $x=-a$ is $x+a$, the two distances are equal. This also places the vertex midway between focus and directrix at $(0,0)$.

For $y^2=20x$, $a=5$, so the focus is $(5,0)$ and the directrix is $x=-5$. The point $(5,10)$ lies on the curve. Its distance from the focus is $10$, and its perpendicular distance from the directrix is also $5-(-5)=10$.

Find a tangent gradient by differentiating the Cartesian equation, then use the negative reciprocal for the normal gradient. A parameter may identify the point, but parametric differentiation is not required.

For the parabola, $2y\,dy/dx=4a$, so $dy/dx=2a/y=1/t$ at $P$. For the hyperbola, $y=c^2x^{-1}$ gives $$\frac{dy}{dx}=-\frac{c^2}{x^2}=-\frac1{t^2}.$$ Insert each point and gradient into $y-y_0=m(x-x_0)$, then simplify.

For $y^2=8x$, $a=2$. At parameter $t=3$, $P=(18,12)$. The tangent is $$3y=x+18,$$ and the normal is $$y=-3x+66.$$ Their gradients $1/3$ and $-3$ multiply to $-1$, checking perpendicularity.

Two matrices can be added or subtracted only when they have the same order. Combine entries that occupy the same row and column; the result has that same order.

For $$A=\begin{pmatrix}2&-1\\3&4\end{pmatrix},\qquad B=\begin{pmatrix}5&2\\-3&1\end{pmatrix},$$ match corresponding positions: $$A+B=\begin{pmatrix}7&1\\0&5\end{pmatrix},\qquad A-B=\begin{pmatrix}-3&-3\\6&3\end{pmatrix}.$$ For example, the lower-left entry of $A-B$ is $3-(-3)=6$.

A scalar is a single number. Multiplying a matrix by a scalar multiplies every entry by that number while leaving the matrix order unchanged.

If $$A=\begin{pmatrix}2&-1\\3&4\end{pmatrix},$$ then $$-3A=\begin{pmatrix}-6&3\\-9&-12\end{pmatrix}.$$ The negative scalar reverses each sign as well as multiplying each magnitude by $3$. Also, $0A$ is the zero matrix of the same order as $A$.

The product $AB$ exists when the number of columns of $A$ equals the number of rows of $B$. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$.

To find one entry, take a row from the first matrix and the matching column from the second, multiply corresponding terms, then add. Repeat for every row-column pair.

Let $$A=\begin{pmatrix}1&2&-1\\0&3&4\end{pmatrix},\qquad B=\begin{pmatrix}2&1\\-1&0\\3&2\end{pmatrix}.$$ Their inner dimensions are both $3$, so $$AB=\begin{pmatrix}1(2)+2(-1)+(-1)(3)&1(1)+2(0)+(-1)(2)\\0(2)+3(-1)+4(3)&0(1)+3(0)+4(2)\end{pmatrix}=\begin{pmatrix}-3&-1\\9&8\end{pmatrix}.$$

The determinant of a 2 by 2 matrix is one number found by multiplying along the main diagonal and subtracting the product along the other diagonal.

For $A=\begin{pmatrix}4&7\\2&3\end{pmatrix}$, $$\det A=4(3)-7(2)=-2,$$ so $A$ is non-singular. By contrast, for $S=\begin{pmatrix}2&6\\1&3\end{pmatrix}$, $$\det S=2(3)-6(1)=0,$$ so $S$ is singular.

An inverse $A^{-1}$ reverses the effect of a square matrix: $AA^{-1}=A^{-1}A=I$. A 2 by 2 inverse exists exactly when the determinant is non-zero.

Keep the main-diagonal entries but swap their positions, change the signs of the other two entries, then multiply the resulting matrix by the reciprocal of the determinant.

For $A=\begin{pmatrix}4&7\\2&3\end{pmatrix}$, $\det A=-2$, so $$A^{-1}=-\frac12\begin{pmatrix}3&-7\\-2&4\end{pmatrix}.$$ Multiplying $A$ by this result gives $I$, which checks both the entry changes and the determinant factor.

The order reverses because $B^{-1}$ must first undo $B$: $$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=I.$$ This relation requires both $A$ and $B$ to be invertible.

A 2 by 2 matrix represents a linear transformation by multiplying each position column vector. The two columns of the matrix are the images of the coordinate basis vectors.

In particular, $\binom{1}{0}\mapsto\binom{a}{c}$ and $\binom{0}{1}\mapsto\binom{b}{d}$. Knowing those two images therefore determines the whole matrix.

If $A=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$, then $$\begin{pmatrix}4\\-2\end{pmatrix}\mapsto\begin{pmatrix}2(4)+1(-2)\\-1(4)+3(-2)\end{pmatrix}=\begin{pmatrix}6\\-10\end{pmatrix}.$$

If $B$ acts first and $A$ acts second, then a vector follows $$\mathbf v\mapsto B\mathbf v\mapsto A(B\mathbf v)=(AB)\mathbf v.$$ Thus $AB$ represents $B$ followed by $A$: the right-hand factor acts first.

A standard transformation matrix is identified by what it does to coordinates, or equivalently by the images of the basis vectors in its columns. All transformations here are centred at the origin where a centre is needed.

A rotation of $210^\circ$ anticlockwise has $\cos210^\circ=-\sqrt3/2$ and $\sin210^\circ=-1/2$, so its matrix is $$\begin{pmatrix}-\frac{\sqrt3}{2}&\frac12\\-\frac12&-\frac{\sqrt3}{2}\end{pmatrix}.$$ Exact trigonometric values keep the transformation exact.

To identify an unfamiliar matrix, test its effect on $(1,0)$ and $(0,1)$ and compare both images with the table. A full description includes the transformation type, line or centre where relevant, direction and angle for a rotation, and scale factor for a stretch or enlargement.

A sequence of linear transformations is represented by a matrix product in reverse reading order: the matrix for the last transformation is placed on the left.

Let $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ rotate $90^\circ$ anticlockwise, and let $S=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ stretch parallel to the $x$-axis by factor $2$. Rotation followed by stretch has matrix $$SR=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&-2\\1&0\end{pmatrix}.$$

For $\mathbf v=\binom{1}{2}$, the separate transformations give $\binom{1}{2}\mapsto\binom{-2}{1}\mapsto\binom{-4}{1}$. The product gives $SR\binom{1}{2}=\binom{-4}{1}$, confirming the order.

Repeating the same transformation $A$ exactly $n$ times gives $A^n$. For example, four successive quarter-turns give $R^4=I$, returning every vector to its original position.

A non-singular transformation can be undone by its inverse matrix. Its determinant also controls area: every finite region has its area multiplied by the absolute value of the determinant.

For $M=\begin{pmatrix}2&1\\0&3\end{pmatrix}$, $\det M=6$. Hence $$M^{-1}=\frac16\begin{pmatrix}3&-1\\0&2\end{pmatrix},$$ and a region of area $5$ is transformed to a region of area $|6|\times5=30$. Conversely, divide the image area by $6$ to recover the original area.

For invertible transformations, $(AB)^{-1}=B^{-1}A^{-1}$: the inverse first undoes the action performed last. This is the transformation meaning of the reverse-order inverse rule.

A finite sum can be evaluated by rewriting its summand as a polynomial in the index, splitting it term by term, and applying the standard results. Here $n$ is a positive integer.

Summation is linear: constants may be taken outside and separate polynomial terms may be summed separately. Thus $$\sum_{r=1}^{n}(ar^3+br^2+cr+d)=a\sum_{r=1}^{n}r^3+b\sum_{r=1}^{n}r^2+c\sum_{r=1}^{n}r+dn.$$ Expand or factor the summand before choosing the formulas.

For the official model form, $$\sum_{r=1}^{n}r(r^2+2)=\sum_{r=1}^{n}(r^3+2r)=\left(\frac{n(n+1)}2\right)^2+n(n+1).$$ Factoring gives $$\frac{n(n+1)}4\bigl(n(n+1)+4\bigr).$$ This preserves the equality for every positive integer $n$, rather than checking only particular values.

For a different lower limit, define $F(m)=\sum_{r=1}^{m}f(r)$. Then $$\sum_{r=a}^{b}f(r)=F(b)-F(a-1).$$ For example, with $f(r)=r^2+2$, $$\sum_{r=4}^{10}(r^2+2)=F(10)-F(3)=(385+20)-(14+6)=385.$$