STEP II 1992 question 10 solution

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

STEP II 1992 question 10 solution

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 10 solution

i) arg(z-1) - arg z = alpha

ii) arg(z-1) - arg z = alpha - arg(-1)

arg(1-z) - arg z = alpha

iii) |z-1| = |z|

iv)

$\displaystyle w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})}$

Shuffle around the negative signs...

$\displaystyle \\w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{1}-z_{4})(z_{3}-z_{2})}\\ \arg (w)=\arg \left(\frac{z_{1}-z_{2}}{z_{1}-z_{4}}\right)+\arg \left(\frac{z_{3}-z_{4}}{z_{3}-z_{2}}\right)\\ \arg (w)=\arg \left(\frac{z_{1}-z_{2}}{z_{1}-z_{4}}\right)-\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{4}}\right)$

Now note that since z1, z2, z3 and z4 lies in that order on a circle in the complex plane, the line joining z2 and z4 is a chord and z1 lies above this chord while z3 lies below this chord. We then have, deducing from the result in part(i) and part(ii):

$\displaystyle \\ \arg \left(\frac{z_{1}-z_{2}}{z_{1}-z_{4}}\right)=\alpha\\ \arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{4}}\right)=\alpha-\pi\\ \arg (w)=\arg \left(\frac{z_{1}-z_{2}}{z_{1}-z_{4}}\right)-\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{4}}\right)\\ \therefore \arg (w)=\pi\\ \implies \Im{\{w\}}=0$