z - 2w* = i ①
iz - w = i ②

Let z = a + bi, and w = c + di, where a,b,c,d are real numbers. Since w* is w's conjugate, w* = c - di

① becomes :
a + bi - 2(c - di) = i
a - 2c + bi + 2di = i
(a - 2c) + (b + 2d)i = 0 + i

Comparing coefficients of the real portion on both sides,
a - 2c = 0 → a = 2c ③

Comparing coefficients of the imaginary portion on both sides,
b + 2d = 1 ④

② becomes :
i(a + bi) - (c + di) = i
ai + bi² - c - di = i
-b - c + (a - d)i = 0 + i

Comparing coefficients of the real portion on both sides,
-b - c = 0 →c = -b ⑤

Comparing coefficients of the imaginary portion on both sides,
a - d = 1 ⑥


Sub ⑤ into ③,
a = -2b

Sub a = -2b into ⑥,
-2b - d = 1
d = -2b - 1

Sub d = -2b - 1 into ④,
b + 2(-2b - 1) = 1
b - 4b - 2 = 1
-3b = 3
b = -1

Then,
d = -2(-1) - 1 = 1
a = 2b = -2(-1) = 2
Sub b = -1 into ⑤, c = -(-1) = 1

Therefore
z = 2 - i
w = 1 + i

bi)

u = -1 - √3 i

|u| = √((-1)² + (-√3)²)

= √(1 + 3)
= √4
= 2


For arg(u) ,

basic angle
= tan-¹ (-√3 / -1)
= tan-¹ √3
= π/3 rad (or 60°)

Since this complex number has both coefficients being negative, it is in the '3rd' quadrant, where the Re and Im portion are negative.

As –π < θ ⩽ π (see syllabus definition for range of the argument),

the argument must be negative and not π + π/3 (as we would do for normal trigo)

So arg (u) = -π + π/3 = -2π/3

bii)


Written in exponential form,

v = √2 e^(7πi/12)

u = 2 e^(-2πi/3)

Now recall that a complex number and its conjugate are 'mirror images' on the complex plane, reflected about the real axis.

They have the same modulus, but their arguments are negatives of each other.

so u* = 2 e^(2πi/3)

(Try to verify that uu* = |u|² with this)




Therefore,


vu* = √2 e^(7πi/12) (2 e^(2πi/3))

= 2√2 e^(7πi/12 + 2πi/3)

= 2√2 (5πi/4)


Since -π < θ ≤ π, we have to write an equivalent value of the argument.

= 2√2 (5πi/4 - 2π)

= 2√2 (-3πi/4)

biii)

For (vu*)ⁿ to be purely imaginary, the real portion must equal 0.


So (2√2 e^(-3πi/4))ⁿ has to have a real portion that is 0.


We know that when the real portion is 0, the complex number's coordinates lie somewhere on the imaginary axis.

This is where the argument is π/2 or -π/2.
And if we shift it forward or backwards by multiples of π radians, the the real portion is still 0.

(i.e odd integer multiples of π/2)


Since (2√2 e^(-3πi/4))ⁿ = (2√2)ⁿ e^i(-3nπ/4),
and (-3nπ/4) = -π/2 (3n/2) ,

we just need the lowest possible positive integer n such that 3n/2 is an odd integer

This gives us an odd integer multiple of π/2.
3n/2 becomes 3 when n = 2.


Lowest n = 2