I dare someone to give me worked solutions for these 5 complex number qs (not hard!) (1 Viewer)

[hscsubjectstodo]

QUESTION 2 (15 Marks) [START A NEW PAGE]

(√3) + i
(a) Given z = ---------- ,
1 + i

(i) Find the argument and modulus of z
(ii) Find the smaller positive integer n such that z^n is real

(b) The complex number z moves such that Im[1 / conj(z) - i] = 2
Show that the locus of z is a circle and find its centre and radius.

(c) Sketch the region in the complex plane where the inequalities
| z + 1 - i | < 2 and 0 < arg(z + 1 - i) < 3π/4 hold simultaneously

(d) Find the three different values of z for which
1 + i
z^3 = ------------ .
√2

(e) By applying De Moivre's theorem and also expanding (cosθ + isinθ)^3,
express cos3θ as a polynomial in cosθ.

 

[hscsubjectstodo]

Re: I dare someone to give me worked solutions for these 5 complex number qs (not har

ALSO NOTE WITH PART A:
Z=
(√3) + i
----------
1 + i

AND PART D:
Z =
1+ i
------
√2

 

[life92]

Re: I dare someone to give me worked solutions for these 5 complex number qs (not har

a)i) (√3 + i) / (1+i)
= 2 (√3/2 + i/2) / [ (1/√2)(1/√2 + i/√2)
= 2√2 (cis pi/6) / cis (pi/4)
= 2√2 cis -pi/12

Mod z = 2√2
Arg z = -pi / 12

ii) So to be real, arg z must = 0 or pi
Therefore, n = 12, to make arg z = -pi, which is the same as pi

b) Im[1 / conj(z) - i] = 2

1 / conj(z) - i = z / (mod z)^2 - i
= (x+iy) / (x^2+y^2) - i
Therefore, Im[1 / conj(z) - i] = y / (x^2 + y^2) - 1
Now, y / (x^2 + y^2) - 1 = 2
y / (x^2 + y^2) = 3
y = 3x^2 + 3y^2
3x^2 + 3y^2 - y = 0
x^2 + y^2 - y/3 = 0
x^2 + (y-1/6)^2 = 1/36
Therefore, centre (0,1/6)
radius = 1/6

c) Don't know how to show you

d) z^3 = 1/√2 + i/√2
= cis pi/4

now z = r cis theta
r^3 cis 3theta = cis pi/4
therefore, r = 1
3 theta = pi/4 , 9pi/4, -7pi/4
theta = pi/12, 3pi/4, -7pi/12

Therefore, z = cis theta where theta = pi/12, 3pi/4, -7pi/12

(e) By applying De Moivre's theorem and also expanding (cosθ + isinθ)^3,
express cos3θ as a polynomial in cosθ.

(cosθ + isinθ)^3 = cos^3 θ + 3 i cos^2 θ sin θ - 3 cosθ sin^2 θ - i sin^3 θ
and (cosθ + isinθ)^3 = cos3θ + isin3θ

Therefore, cos3θ + isin3θ = cos^3 θ + 3 i cos^2 θ sin θ - 3 cosθ sin^2 θ - i sin^3 θ
Equating real parts:
cos3θ = cos^3 θ - 3 cosθ sin^2 θ
= cos^3 θ - 3 cosθ (1-cos^2 θ)
= cos^3 θ - 3 cos θ + 3 cos^3 θ
= 4 cos^3 θ - 3 cos θ

So cos3θ = 4 cos^3 θ - 3 cos θ

Hopefully that helps !