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# P6-Panic Time- Complex nos. watch

1. i seem to be asking a lot of questions,
but its only questions that i have
and answers short of.

i have appended my p6 problems. all of ur help wud be appreciated . thank you
have a great day.
Attached Images
2. P6 complex nos..pdf (52.1 KB, 147 views)
3. 70

Let z = x+iy. Then

Then Im[(2z+1)/(iz+1)]

= Im[(2x+1+2iy)/(1-y+ix)]

= Im[(2x+1+2iy)(1-y-ix)/((1-y)^2+x^2)] <multiply by conjugate of the denominator>

= [2y(1-y) - x(2x+1)]/[1-2y+y^2+x^2]

So Im[...] = -2 if

2y - 2y^2 - 2x^2 - x = -2 +4y - 2y^2 - 2x^2

2 = x +2y

Which is the equation of a line
4. 72

As |z-3| = 2 is a circle, centre 3, radius 2, then |z|=5 is the maximum, attained at z = 5
5. Q48(Page2)

z describes y=1 so z=a+i.

If
This is, of course, a parabola of the form but translated one unit in the direction of the negative x axis.
6. 112

(a) The three cube roots of -64 are

-4

4 (cos(pi/3) + i sin(pi/3))

4 (cos(2pi/3) + i sin(2pi/3)

and they sit equally spaced around the circle |z|=4

(b) The sixth roots of -64 are

2 [cos ((2n+1) pi/6) + i sin((2n+1)pi/6)]

for n = 0, 1, 2, 3, 4, 5

and they sit equally spaced around the circle |z|=2
7. 34

The map W = 2(Z+ 3 - i)

maps the unit circle to the circle with centre 3-i and radius 2

If w = (1-z)/z maps the line y = 1/2 to the unit circle

Then the combination of the two will map the line y = 1/2 to this second circle - and this equals

(- 2Z - 5 + 2i)/(2Z + 6 - 2i)
8. sin7t = 7 sint

if (from the formular for sint7t)

-64 (sint)^7 + 112 (sint)^5 - 56 (sint)^3 = 0 (*)

If sint =/= 0 then

-64 (sint)^4 + 112 (sint)^2 - 56 =0

Write z = (sint)^2

-64 z^2 + 112 z - 56.

8z^2 - 14z + 7 = 0

Now this has discriminatnt

14^2 - 4.7.8 = -28

and so it has no real roots.

So the equation (*) has real roots for sint only when sint=0 i.e. when t = n pi
9. w = (1+3z)/(1-z)

Find the image of Rez = 0

Method 1: a general point on the line is iy

(1+3iy)/(1-iy) = (1+3iy)(1+iy)/(1+y^2)

So

u = (1 -3y^2)/(1+y^2) = 1 - 4y^2/(1+y^2)
v = (4y)/(1+y^2)

Then (1-u)/v = y

Substituting into the v-equation:

v [1 + (1-u)^2/v^2] = 4(1-u)/v

v^2 + 1 - 2u + u^2 = 4 - 4u

u^2 +2u +v^2 = 3

(u+1)^2 + v^2 = 4

Method 2:

(Let z* denote z conjugate in the following - note for complex numbers (ab)* = a*b* and (a/b)* = a*/b* and |a|^2 = aa*)

w = (1+3z)/(1-z)

w - wz = 1 + 3z

z(3+w) = w - 1

z = (w-1)/(3+w)

Now w is on the image C only when z is on the imaginary axis - this has equation z = - z*

So w is on C precisely when

(w-1)/(3+w) = - [(w-1)/(3+w)]* = - (w*-1)/(3+w*)

So

(w-1)(3+w*) = - (w*-1)(3+w)

-3 + 3w - w* + ww* = 3 - 3w* + w - ww*

ww* +w* + w - 3 = 0

(w+1)(w*+1) = 4

|w+1|^2 = 4

|w+1| = 2

as required
10. (Original post by RichE)
70

Let z = x+iy. Then

Then Im[(2z+1)/(iz+1)]

= Im[(2x+1+2iy)/(1-y+ix)]

= Im[(2x+1+2iy)(1-y-ix)/((1-y)^2+x^2)] <multiply by conjugate of the denominator>

= [2y(1-y) - x(2x+1)]/[1-2y+y^2+x^2]

So Im[...] = -2 if

2y - 2y^2 - 2x^2 - x = -2 +4y - 2y^2 - 2x^2

2 = x +2y

Which is the equation of a line
oh thank you thank you thank you thank you...for all your help. u saved the day. hurray
11. (Original post by Gaz031)
Q48(Page2)

z describes y=1 so z=a+i.

If
This is, of course, a parabola of the form but translated one unit in the direction of the negative x axis.
thank you thank you.....

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