A few complex number questions (1 Viewer)

[x.Exhaust.x]

1a) Find all integers x and y such that (x+iy)^2=5-12i

I got x=± 3, y= ∓2 in the end. Is that correct?

b) Hence, solve the equation z^2-(5+4i)z+1+13i=0 for z

2. Find the complex square roots of 16i

3. Given that z = cis2theta-1/cis2theta+1, express z in Cartesian form

b) Hence evaluate:

i) Re(z)
ii) Im(z)
iii) | z |

c) Write down the simplest expression for | w | given that w = z + 1

| w | = z-1? Or am I confusing myself with the dash of w at the top?

d) Suggest any possible values of Arg(z)

4. Consider the monic quadratic equation y= ax^2 + bx + c such that b^2-4ac<0. Given that the only turning point of this curve has the coordinates (M, N), express the roots of the quadratic equation ax^2+ bx + c in terms of M and N.

5. Given that two complex numbers have non-zero imaginary parts and that their sum and product are both real numbers, show that these two complex numbers must be complex conjugates of each other.

 

[Just.Snaz]

omg, 4 unit maths, how i've missed you <3

yes a) is correct. b) you use the quadratic formula and then you use part a) for under the square root. You should get 5-12i under the square root.

for 2. let sqrt 16i = a + ib
so (a + ib)^2 = 16, and just do the same as 1.

3. expand cis form and then multiply the bottom by (cos@ + 1) - isin@. always group the real and imaginary together then rationalise the bottom by x by the conjugate.

for b) Re(z) means you only take the real part. Imaginary numbers are always in the form of a + ib. a is the real part, b is the imaginary part. - even though the actual b value is real. So for Im(z) you take b.
z = sqrt a^2 + b^2.
vital things you must know.

I would do the rest but I have chem exam this thursday so if no one else does them, i'll be back in my break anyway

 

[x.Exhaust.x]

Thanks. All the best with your chem exam

Edit: Need solutions to the above q's.

 

[Zeber]

5. Given that two complex numbers have non-zero imaginary parts and that their sum and product are both real numbers, show that these two complex numbers must be complex conjugates of each other.

let
z = x+iy
k = a+ib where a,x,y and b are real and y and b =/= 0.

z+k, we have x+a+i(y+b), but since z+k is real, y+b = 0 and .'. y = -b or b = -y

zk = ax + ixb + aiy - yb
that is, xb + ay = 0 as zk is real.

from y = -b
xb - ab = 0
b(x-a) = 0
x-a = 0
x = a

Gathering it all together, you can see that the two complex numbers must be conjugates

 

[gurmies]

Nice Zeber, beat me to it =) Now too sure about |w| question. Don't confuse that with conjugate of w (line on top). These "| |" are the notation for modulus. Since w = z + 1, where z = x + iy (any complex number), w = (x+1) + iy. |w| = root ((x+1)^2 +y^2). Not sure how to further simplify that or what a possible argument could be?

 

[Zeber]

|w| = sqrt[(x+1)^2 + y^2] when you let w = x+iy

argument should be -pi < x < pi as i don't see the +1 having an effect on the overall domains of the argument.

 

[gurmies]

oh, so that simple?

 

[Zeber]

i'm not 100% sure, but if you think about it.

if z = x+iy, it has all the arguments, I don't see an effect of adding 1 to it changing the overall domain of the arguments?

 

[gurmies]

Wouldn't the vector "z" just be shifted one unit to the left? Therefore there is no change in modular length or principle argument.

 

[Zeber]

nah i think the vector z would have a greater magnitude but no change in argument (think of 1+i, if it becomes 2+i, the modulus changes)

 

[gurmies]

yes, you're completely correct! My mistake:burn: . There's definitely no change in principal argument however

 

[x.Exhaust.x]

More q's!:

1) Without using the binomial theorem, show that (-1+root3i)^n - (-1-root3i)^n has no real part.

2) Prove
______ ___ ___
(z1/z2) = z1/ z2

3) Given that z=cistheta, find the value of alpha such that (1+root3i)z=2cis(theta + alpha)

4a) Find all distinct values of z for which z^5 = 16root2(1+i)

5) Find the roots to the equation z^4 = 1 through De-moivre's theorem

6) Express 3 + 4i in mod-arg form (found r=5, finding argz or tantheta is bugging me)

7) Evaluate Im(1+cosntheta + isinntheta) / (1+cosntheta - isinntheta)

8) Express the square roots of 3+ 4i in the form a+ib where a and b are real numbers

Thanks.

 

[bored of sc]

More q's!:

1) Without using the binomial theorem, show that (-1+root3i)^n - (-1-root3i)^n has no real part.

By induction:

i) Show for n = 1.
(-1+rt3i)¹-(-1-rt3i)¹ = -1+rt3i+1+rt3i = 2rt3i = purely imaginary i.e. real part = 0

ii) Assume true for n=k
(-1+rt3i)^k-(-1-rt3i)^k = Mi where M is the coefficient of the complex number

iii) Prove true for n=k+1
(-1+rt3i)^k+1-(-1-rt3i)^k+1
= (-1+rt3)^k*(-1+rt3i)-(-1-rt3i)^k*(-1-rt3i)
= [Mi+(-1-rt3i)^k](-1+rt3i)-(-1-rt3i)^k(-1-rt3i) ---- from (ii)
= Mi(-1+rt3i)+(-1-rt3i)^k(-1+rt3i)-(-1-rt3i)^k(-1-rt3i)
= -Mi-Mrt3+(-1-rt3i)^k(-1+rt3i-(-1-rt3i))
= -Mi-Mrt3+(-1-rt3i)^k*2rt3i ---- from (i)
= -M(i+rt3)+(-1-rt3i)^k*2rt3i
= confused

Damn. Induction probably isn't the way to go.

 

[bored of sc]

3) Given that z=cistheta, find the value of alpha such that (1+root3i)z
= 2cis(theta + alpha) ---- (1)

For 1+rt3i

r = rt[1²+(rt3)²] = rt(1+3) = rt4 = 2

tan(alpha) = rt3/1
alpha = tan^-1(rt3/1) = 60^o

therefore 1+rt3i = 2cis60

(1+rt3i)z = (2cis60)(cistheta) = 2cis(theta+60) ---- (2) ---- laws of multiply complex numbers i.e. multiply moduli and add arguments

thus, alpha = 60^o ---- equating (1) and (2)

 

[bored of sc]

6) Express 3 + 4i in mod-arg form (found r=5, finding argz or tantheta is bugging me)

r = rt(3²+4²) = rt(9+16) = rt25 = 5
theta = tan^-1(4/3) = 53.13010235^o = 53^o (to nearest degree)

so 3+4i = 5cis53 in mod-arg form

 

[bored of sc]

8) Express the square roots of 3+ 4i in the form a+ib where a and b are real numbers

3+4i = (a+ib)² = a²+2iab-b²

now we equate real and imaginary parts:
therefore 3 = a²-b² --- (1)
4 = 2ab --- (2)

from (2)
b = 2/a (3)

sub into (1)

3 = a²-(2/a)²
3 = a² - 4/a²
3a² = a⁴-4
a⁴-3a²-4 = 0
(a²-4)(a²+1) = 0
but a is real so only take a²-4

thus a = +2

sub into (3)
b = +1

thus, sqrt(3+4i) = +2+i ---- note: it must be (2+i) and (-2-i)

 

[Trebla]

1) (-1 + i√3)ⁿ - (-1 - i√3)ⁿ
= 2ⁿ[cis (2π/3)]ⁿ - 2ⁿ[cis (-2π/3)]ⁿ
= 2ⁿ{cis (2πn/3) - cis (-2πn/3)}
= 2ⁿ{cos (2πn/3) + isin cos (2πn/3) - cos (-2πn/3) - isin (-2πn/3)}
But cos (-x) = cos x and sin (-x) = - sin x
= 2ⁿ{cos (2πn/3) + isin cos (2πn/3) - cos (2πn/3) + isin (2πn/3)}
= 2ⁿ[2isin (2πn/3)]
which is purely imaginary (no real part)

 

[gurmies]

______ ___ ___
(z1/z2) = z1/ z2

Okie dokie, let z1 = a + ib
z2 = x + iy

z1/z2 = (a+ib)/(x+iy)

= (a+ib)/(x+iy) x (x-iy)/(x-iy)

= (a+ib)(x-iy)/(x^2+y^2)

= (ax + by - i(ay -bx))/(x^2 + y^2)


_____
(z1/z2) = (ax + by + i(ay - bx))/(x^2+y^2)

= (a-ib)(x+iy)/(x^2 + y^2)

= (a-ib)/(x-iy) x (x+iy)/(x+iy)

= (a-ib)/(x-iy)
__ __
= z1/ z2

Next question, z^4 = 1.

All have modulus of 1, with angular spacing of pi/2 around the unit circle. Thus the roots are: cis 0, cis pi/2, cis pi, cis (-pi/2). In other words, roots are: 1, i, -1, -i respectively