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"The Level 1 and 2 Mathematics and Statistics courses are compulsory (bit see below on MATH2931) and the others are strongly recommended. MATH1151/1251 replace MATH1131/1231 in the formal requirements for a Mathematics or Statistics major. MATH2901 and MATH2931 are required for Part I exemption...

check the question... should it be (x+1)^2 instead of (x^2+1)^2 Btw is x a real number? If so.... think modulus

ACCT1501 - Prescribed Textbook only... dunno about practice set. ECON1101 - Prescribed Textbook only MATH1151 - Don't buy Salas. It's not needed in first year nor second year maths. The course notes (not shown on textlist but is required) are sufficient. Dunno about MGMT

Um not quite right Z=1/(z-3)+17/3 does not imply |Z|=|1/(z-3)|+17/3.... think triangle inequality... rather Z-17/3=1/(z-3) --->|Z-17/3|=1/3 Explanation: Instead of considering z, consider the complex number m=1/(z-3). |z-3|=3 implies |1/(z-3)|=1/3, which implies |m|=1/3, so it means...

actually i think you are right but you just got confused The locus of 1/(z-3) was a circle of same centre so 1/(z-3) + 17/3 equals to a locus of (x-17/3)^2+y^2=1/9. (51/9=17/3) This is because every point 1/(z-3) on the circle shifts right by 17/3, so logically the centre of the circle should...

Close but not quite.... cos adding 3+i onto the vector 2z should move it upwards by i and right by 3, so logically the centre of the circle should shift by the same amount Or convert into parametrics question: Let P=z=x+iy so x^2+y^2=k^2 (x,y,X,Y are real) Let Q=X+iY such that Q=2z+3+i...

This part requires you assume z=cosx+isinx which assumes |z|=1 Rather: ((z+1)/(z-1))^n=-1 Taking mod from both sides gives: |z+1|^n/|z-1|^n=1 |z+1|/|z-1|=1 |z+1|=|z-1| Solving for z geometrically implies z is purely imaginary or 0, and hence can be expressed as z=ia where a is real.

It is perfectly clear what OP is asking for. The graph can be rearranged into a hyperbola: x^2 - 4y^2 = 4 --> x^2/4 - y^2 = 1 noting asymptotes of a hyperbola (of the form x^2/a^2-y^2/b^2=1) is y=bx/a and y=-bx/a then the asymptotes are y=x/2 and y=-x/2.

arg(2)=0 btw..... so that doesnt work (division via 0) |z-2|=2 doesnt imply z-2=2 or -2.... z-2 could be 1+sqrt(3)i An algebraic solution would be |z-2|=2 --> z-2=2cisx for some x (ie arg(z-2)=x) z=2(cosx+1+isinx) Note cosx+1=2cos^2(x/2) and sinx=2cos(x/2)sin(x/2) z=2(2cos(x/2))cis(x/2) after...

I do actuarial/maths/stats: Non actuarial commerce courses (aka acct+econ): Just do homework.... rest can be crammed b4 assessments. So on avg around 1-2 hours Actuarial commerce courses: Homework+readings+re-reading lecture notes until material is completely understood.... so avg 4-5...

Lmao discrete maths with 4 lectures on 4 days... happens every year

Not at UNSW.

I think you got the question wrong. It should be (1-cosx-isinx)/(1+cosx+isinx) (test x=pi/2 to see the original is incorrect) So you should get: tan(x/2)(sin(x/2)-icos(x/2))/(cos(x/2)+isin(x/2)) =tan(x/2)(cos(pi/2-x/2)-isin(pi/2-x/2))/cis(x/2) noting cos(x)=cos(-x) and -sin(x)=sin(-x)...

w^3=1 ==> (w-1)(w^2+w+1)=0 But w is not real so w=/=1 so w^2+w+1=0 and hence w=-w^2-1

Modulus method (a+ib)^2=-12-16i |(a+ib)^2|=|-12-16i| noting |z^2|=|z|^2 |a+ib|^2=sqrt(144+256) noting |a+ib|=sqrt(a^2+b^2) a^2+b^2=20 Then combine with a^2-b^2 to get the answer

x^2=(16-8t^2+t^4)/(4+t^2)^2 y^2=16t^2/(4+t^2)^2 x^2+y^2=(16-8t^2+t^4)/(4+t^2)^2+16t^2/(4+t^2)^2 = (16+8t^2+t^4)/(4+t^2)^2 = (4+t^2)^2/(4+t^2)^2=1

Make the denominator real: z=(2+it)^2/[(2+it)(2-it)]=(4-t^2+4it)/(4+t^2) Let z=x+iy x=Re(z)=(4-t^2)/(4+t^2) y=Im(z)=4t/(4+t^2) The rest just involves some parametrics work in 3u to get rid of t (hint square x and y and then combine) You should get x^2+y^2=1 i think

The vector z-a is the vector from the complex number a to the complex number z IN THE DIRECTION TOWARDS a. The problem with the solutions is that they mistaken the direction of the vector. z-i is going FROM i TO P, while the argument they have shown is for the vector going from P to i, which...

I agree. A more rigorous solution would be: arg(z-i) - arg(z+i) = pi/2 let z=x+iy , x,y are real arg((z-i)/(z+i))=pi/2 arg((x+iy-i)/(x+iy+i)=pi/2 arg((x^2+y^2-1-2ix)/denominator (dont care))=pi/2 ==> real part =0 and imaginary part >0 we also note the denominator (after making it real)...

arctan((-2x)/(x^2 + y^2 -1)) = pi/2 ==> denominator=0 AND numerator is positive as tan(pi/2) is +infinity So -2x>0 and x^2+y^2=1 So x<0 and x^2+y^2=1