Complex numbers questions (1 Viewer)

[YBK]

Hey, just a few quick questions

From Cambridge 2.2


4) Use the properties of modulus and argument of a complex number to deduce that:

a) complex conjugate (z1.z2) = complex conjugate z1 . complex conjugate z2
(i'll probably figure out the rest if I understand this one)


8) z=1+root(3)i. Find the smallest positive integer n for which z^n is real and evaluate z^n for this value of n. Show that there is no integral calue of n for which z^n is imaginary.

I got n=0 first, but then read the restriction, so the answer is n=3.

And I'm not sure about the proof.


9) z has modulus r and argument @. Find in terms of r and @ the modulus and one argument of:

a) z^2
b) 1/z
c)iz

Ans:
a) modulus: r^2
argument: 2@
b) modulus: 1/r
argument: -@
c) modulus: r
argument: @.i

Not sure if my answer is right for this one; the book doesnt give a solution to the question :S


Thanks everyone!!!

 

[Sober]

a) complex conjugate (z1.z2) = complex conjugate z1 . complex conjugate z2
(i'll probably figure out the rest if I understand this one)

I will use underline to denote complex conjugate.

LHS = z1 • z2 = (x1 + iy1)(x2 + iy2) = x1x2 - y1y2 + i(x1y2 + x2y1) = x1x2 - y1y2 - i(x1y2 + x2y1)

RHS = z1 • z2 = (x1 - iy1)(x2 - iy2) = x1x2 - y1y2 - i(x1y2 + x2y1) = LHS

Now you can take a stab at the rest.

 

[pLuvia]

4)

Let z₁=a+ib
Let z₂=x+iy

z₁.z₂ = (ax-by)+i(ay+bx)

Complex conjugate of z₁.z₂ = (ax-by)-i(ay+bx)
Complex conjugate of z₁ and z₂ = (ax-by)-i(ay+bx)

QED (can't exactly show the bar at the top)

8)

z=1+√3i
z=2cis(pi/3)
zⁿ=2ⁿ(n(pi/3)

For zⁿ to be real arg = 0,pi,2pi, etc...

.: when (n(pi/3) = 0 n=0
when (n(pi/3) = pi n=3
when (n(pi/3) = 2pi n=6

.: the smallest possible +'ve integer for n is 3 to be real

For zⁿ to be Im arg = pi/2, 3pi/2, 5pi/2

when npi/3 = pi/2 n = 3/2
when npi/3 = 3pi/2 n = 9/2
when npi/3 = 5pi/2 n = 15/2

As seen here for zⁿ to be Im, the n value is not an integer

9)
a) Let z=rcis @
z² = r² cis 2@
Hence mod = r²
arg = 2@

b)
1/rcis @ = (rcis @)^-1 = 1/r cis(-@)

mod = 1/r
arg = -@

 

[Riviet]

For 4, if it's not a sum or subtraction, try letting z=rcis@, it seems to be easier when doing those types of proofs.

 

[YBK]

Thanks! I can do question 1 the way Sober and Pluvia did it, but i'm not sure about Riv's method. I thought that we had to "Use the properties of modulus and argument of a complex number to deduce ", which is why I wasn't sure if the method was acceptable.


question 8, I also did it in a similar way to Pluvia, but has that actually proven that for all z^n to be imaginary, the n value is not an integer?


and Riviet, for question 9c, are you sure we can denote i as cis pie/2? I know that multiplication by i is supposed to rotate anticlockwise by 90 degrees, which is why I put arg as @.i. But to think of it, @ + pie/2 makes more sense. I'm just unsure of the first step you did.


Thanks again

 

[Riviet]

Thanks! I can do question 1 the way Sober and Pluvia did it, but i'm not sure about Riv's method. I thought that we had to "Use the properties of modulus and argument of a complex number to deduce ", which is why I wasn't sure if the method was acceptable.

Sorry, forgot about that condition asked in the question, but questions involving proofs just ask you to prove some identity, which I recommend you tryout using rcis@ instead of z=x+iy.

and Riviet, for question 9c, are you sure we can denote i as cis pie/2? I know that multiplication by i is supposed to rotate anticlockwise by 90 degrees, which is why I put arg as @.i. But to think of it, @ + pie/2 makes more sense. I'm just unsure of the first step you did.


Thanks again

Yes, I am sure you can use cis(pi/2), because if you simplify cos(pi/2) + isin(pi/2), it is equal to 1. Also remember that when you multiply by i, you are ADDING pi/2 to the argument, not multiplying it.

 

[pLuvia]

If you were using modulus and argument to prove z₁.z₂ then

Let z₁ = r(cos@+isin@)
Let z₂ = a(cos#+isin#)

z₁.z₂ = ar(cos[@+#]+isin[@+#])

Complex conj of z₁.z₂ = ar(cos[@+#]-isin[@+#]
Complex conj of z₁ and z₂ = ar(cos[@+#]-isin[@+#] {Just multiplying the conjugates of z1 and z2 you should obtain that answer}

QED

 

[YBK]

Sorry, forgot about that condition asked in the question, but questions involving proofs just ask you to prove some identity, which I recommend you tryout using rcis@ instead of z=x+iy.

Yes, I am sure you can use cis(pi/2), because if you simplify cos(pi/2) + isin(pi/2), it is equal to 1. Also remember that when you multiply by i, you are ADDING pi/2 to the argument, not multiplying it.

Ahh! That makes sense now stupid me, no idea why i multiplied argument by i .

 

[YBK]

If you were using modulus and argument to prove z₁.z₂ then

Let z₁ = r(cos@+isin@)
Let z₂ = a(cos#+isin#)

z₁.z₂ = ar(cos[@+#]+isin[@+#])

Complex conj of z₁.z₂ = ar(cos[@+#]-isin[@+#]
Complex conj of z₁ and z₂ = ar(cos[@+#]-isin[@+#] {Just multiplying the conjugates of z1 and z2 you should obtain that answer}

QED

Awsome! That answers question


thanks all

 

[pLuvia]

No problem