Difficult complex questions, help!! (1 Viewer)

[carolinex]

1. z1 and z2 are two complex numbers such that (z1 + z2)/(z1 - z2) = 2i.
a) On an Argand diagram show vectors representing z1, z2, z1+z2 and z1-z2 [i dont think you can draw diagrams here, but could you describe it or something?]
b) Show that |z1| = |z2|
c) If x is the angle between the vectors representing z1 and z2 show that tan(x/2) = 1/2
d) Show that z2 = 1/5 (3+4i)z1

2.
a) Write down the modulus and argument of each i and -i
b) Show on the unit circle on an Argand Diagram the two square roots of i(z1 and z2) and the two square roots of -i(z3 and z4)
c) P(x) = x^4 + 1. Show that the roots of P(x)=0 are z1, z2, z3 and z4, and factor P(x) completely over the real numbers.

[from CSSA 1993]
thank you

 

[tohriffic]

Hi do you have the answers to these questions?

Regarding question 1a), have they arbitrarily plotted the points or do they have specific numeric values? D:

If it is arbitrary,

This is just a case of simple vector addition and subtraction. Just plotting z1 and z2 randomly mind you.

 

[seanieg89]

1.
The condition given implies that the parallelogram spanned by and is a rhombus (perpendicular diagonals) with the diagonal having length twice that of the diagonal . (Also make sure the orientation of your rhombus is correct.)
c) is simple trig once you understand the geometry of the situation.
d) Follows from c), and from how we rotate complex numbers.

2.

End factorisation is by pairing complex conjugates or directly by using a difference of squares trick.

You should fill in the details I glossed over as an exercise.

 

[tohriffic]

Adding to what Seanieg said (for 1b), you find the argument of the given condition. Using your knowledge of args, you will be able to rearrange yourself and (hint) find yourself a certain property of a quadrilateral allowing you to determine what it is.

 

[SpiralFlex]

Adding to what Seanieg said (for 1b), you find the argument of the given condition. Using your knowledge of args, you will be able to rearrange yourself and (hint) find yourself a certain property of a quadrilateral allowing you to determine what it is.

Faster way than that.

 

[seanieg89]

Faster way than that.

multiplying out i'm guessing, its not like the difference in speed is substantial...

 

[SpiralFlex]

It's my preference though.

 

[tohriffic]

Faster way than that.

Please enlighten me.

 

[SpiralFlex]

Please enlighten me.

3k posts! YAY!

 

[tohriffic]

3k posts! YAY!

Haha gosh I didn't even think of doing that, I just went straight to the args. Guess I just like /arguments/ more.

That works too. Ty for that. And congrats on your post count.

 

[math man]

Referring to diagram above let M be midpoint of two diagonals.....Now spiral already proved part b the most efficient way which everyone should do...so for part c angle CAD = x/2 as diagonals of a rhombus bisect the vertice angles. angle AMC is 90 as diagonals of rhombus bisect at 90. Now as diagonals bisect each other, MC = (1/2)|z1 - z2| and AM = (1/2)|z1 + z2|. So in the right triangle AMC tan(x/2) =
|z1 - z2|/|z1+z2| and this gives tan(x/2) = |(z1-z2)/(z1+z2)|= |1/2i| = 1/2 using the given info.....
For d) you should note that z2= cis(x) z1 by rotation of vectors...so z2 = (cosx + i sinx) z1.....using our right angled triangle for tan(x/2) = 1/2 and then half angle formulas you can show that cosx= 3/5 and
sinx = 4/5... therefore z2 = (3/5 + i4/5) z1

 

[carolinex]

3k posts! YAY!

sorry, i dont understand this!! ><"

 

[AAEldar]

sorry, i dont understand this!! ><"

First line to second is just expanding and then regrouping the terms.

Third line is taking the absolute value of both sides (nothing is worked out yet).

Fourth line is the equivalence - the absolute value (or the modulus) of (1-2i) is the same as (1+2i) hence you can get rid of them so to speak.

 

[carolinex]

First line to second is just expanding and then regrouping the terms.

Third line is taking the absolute value of both sides (nothing is worked out yet).

Fourth line is the equivalence - the absolute value (or the modulus) of (1-2i) is the same as (1+2i) hence you can get rid of them so to speak.

oh right. cool thanks