STEP Maths I, II, III 1993 Solutions

I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

[edit] Here it is anyway.
$\int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx$
By the product/sum formulae that no one remembers.
$= ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}$

But if m=±n, one of the fractions explodes. So in that case the question is:

$\int \cos^2 {mx} dx$

$=½ \int 1+ cos{2mx}dx$

$= ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}$

$=\pi$

If m=n=0, the integral turns to $2\pi$.

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt

$I = \int \sqrt{\frac{x+1}{x}}dx$

$= \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt$

$= \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt$

$= \int 2 \cosh^2 t dt$

$= \int 1+ \cosh 2t dt = t + ½ \sinh 2t$ Ignoring the +c for now

$= t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c$
Which you can rewrite using the log form of arsinh.

I'm not the greatest person at getting these integration problems out (actually, right now, I suck pretty badly at integration), but there's a reasonable argument that you don't need to worry too much about having sqrt(x) in your working as long as it disappears in the final answer. Which is the case, according to Mathematica.

Silly question, but does the "whole hog" substitution $u=\sqrt{1+1/x}$ work any better?

STEP III Question 6

.......

The next part is bugging me here,

Determine the positions of P for which Q1 and Q2 coincide.

I can see how the points Q1 and Q2 could coincide they are on the same locus. However the positions of p ? Am i suppose to find where Q1 and Q2 coincide and use one of the formulas given to find where it transforms onto the Z plane?

Ive proved a formula i have not yet used, some how i think this has got to come in here.

You made very heavy weather of the first bit. $|a+ib|^2=a^2+b^2 =(a+ib)(a+ib)^*$. So



2nd bit is fine.

3rd bit: Putting k = i, r=1 we have $zz^*-zk^*-z^*k+kk^* = r^2$.

Since $w_2=z^*,w_2^* = z$ we get:$w_2^*w_2-w_2^*k^*-w_2k+kk^* = r^2$. Set $k_2 = k^*$ and we get $w_2^*w_2-w_2^*k_2-w_2k_2^*+k_2^*k_2 = r^2$. So w_2 has equation of circle radius 1, center $k_2 = k^* = -i$.

(Or your reflection in real axis argument is also fine. I just thought I'd show how you can use the first bit).

I'm not seeing an easy solution to the last part. I'd probably parameterize z as $\cos \theta + (1+\sin \theta)i$, find a formula for w and solve for $w=z^*$. (Possibly just writing z=(x+iy) and taking a similar approach will be easier).

STEP III Question 6

A complex number z lying on a circle centre K and radius where K represents the complex number k.

$|z - k | = r^2$

$|x + iy - (a+bi)| = r^2$

$|x - a + i(y-b)| = r^2$

$(x - a)^2 + (y - b)^2 = r^2$
$x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2$

Consider zz*

$(x + iy)(x - iy) = x^2 + y^2$

kz*

$(a + bi)(x - iy) = ax -aiy + bix + by$

k*z

$(a - bi)(x + iy) = ax + aiy - bix + by$

kk*

-k*z - kk* = ax + aiy - bix - by - ax - aiy + bix - by = -2ax - 2by

$(a - bi)(a + bi) = a^2 + b^2$

$x^2 - 2ax + a^2 + y^2 - 2yb + b^2 = r^2$

$(x^2 + y^2) -2ax - 2by + (a^2 + b^2) = r^2$

$zz^* - k^*z - kk^* + kk^* - r^2 = 0$

The locus of P is which represents the complex number z

| z - i | = 1

$w_1 = \frac{z}{z-1}$

$w_1z - w_1 = z$

$z = \frac{w_1}{w_1 - 1}$

$| z - i | = | \frac{w_1}{w_1 - 1} - i |$

$1 = \frac{|w_1 - i(w_1 - 1)|}{|w_1-1|}$

$|w_1 - 1| = |w_1 - iw_1 + i |$

let w_1 = x + iy

$|x + iy - 1| = x + iy -i(x + iy) + i|$

$\sqrt{(x-1)^2 + y^2} = |x + y + iy - ix + i |$

$\sqrt{(x-1)^2 + y^2} = \sqrt{(x+y)^2 + (y - x + 1)^2}$

$(x-1)^2 + y^2 = (x+y)^2 + (y - x + 1)^2$

$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2xy + y^2 - xy + y - xy + x^2 - x + y - x + 1$

$x^2 - 2x + 1 + y^2 = 2x^2 + 2y^2 + 2y - 2x + 1$

$x^2 + y^2 + 2y = 0$

$x^2 + (y+1)^2 - 1 = 0$

$x^2 + (y+1)^2 = 1$

The locus of W_1 is a circle with centre (0,-i) and radius 1 on the argand diagram.

Thus the locus L is

|w_1 + i | = 1

Now this is where it gets tricky.

w_2 = z*

This formula i proved is screaming at me.

$zz* - k*z - kk* + kk* - r^2 = 0$

We know it has to be some circle and you can guess as its going to be of radius one if its the conjugate of z.

Im tempted to just say if locus of z is |z - i | = 1

therefore the locus of the conjugate of z is |z* - i | = 1

this incidentally happens to come out as the locus of L.

Just had another thought. What is the conjugate of a complex number?, as far as im aware its a reflection in the x axis of the original complex number on the argand diagram. So if the locus of Z is a circle, if you take all the specific points you could make its conjugates by reflecting in the x axis.

So if the locus of Z is a circle centered at (0,i) and has a radius of 1 it touches(0,0) a reflection of this is simply a circle centered (0,-i) with radius 1.

This happens to be x^2 + (y + 1)^2 = 1

Which is the same locus as L.

The next part is bugging me here,

Determine the positions of P for which Q1 and Q2 coincide.

I can see how the points Q1 and Q2 could coincide they are on the same locus. However the positions of p ? Am i suppose to find where Q1 and Q2 coincide and use one of the formulas given to find where it transforms onto the Z plane?

Ive proved a formula i have not yet used, some how i think this has got to come in here.

we know w_2= -z/(z-i) and we also know w_1=z/(z-1)
we just let w_1=w_2
then we can find z=1+i (if my calculation is right lol)

I had a crack at this again, god it is so weird to have done that question 2 years ago nearly. Anyways without looking at what you told me, as i said 2 years ago, the equation i proved in the first part was screaming at me and indeed for good reason, why I did not click no idea because with it, the 3rd part came out fine.

I got w_2 = -iz/(z - i) btw.

Then the last part just fell out really. I might write it up later on, unless you want to do it ?

Rounding down at the end for obvious reasons. Now, I think I missed a joke here: what do LEP and VOZ stand for?

STEP II, Question 16

Attached is my solution. It's a bit ugly, so improvements are welcome.

-$\delta\nu$

They are both simple Caesar encryptions of "IBM".

I J K L M N O P Q R S T U V
B C D E F G H I J K L M N O
M N O P Q R S T U V W X Y Z

So I'm guessing that's the joke. (There's a long running rumour that the name of the computer HAL from "2001: A Space Odyssey" was a reference to IBM; the names here are simply taking the joke further).