Complex numbers help!! (1 Viewer)

[wiliaamnguuyen]

Hey guys! It would really be great if you guys helped me out with a few questions that I can't do. If you're helping me, please show working out so I know how to get there!
Thanks!

The questions are :
1. If z is a complex number where neither Re(z) or Im(z) is zero and it is given that z+1/z is real, find [z] (absolute values bars)

Thanks guys!

 

[lpodtouch]

Hi there!
Wow, you're very keen considering how you're in year 10 yet doing 4u. That's always good to hear.

For the first question, z+ (1/z), change z= a + ib

a + ib + (1/(a+ib)
Multiply 1/(a+ib) with its conjugate to receive:
= a+ib + [(a-ib)/a^2+b^2)]
Afterwards, multiply a+ib with a^2 + b^2 to form a common denominator:
= [(a+ib)(a^2+b^2)+a+ib]/(a^2+b^2)
Expand these parentheses to form:
= [a^3 + ab^2 + a + i(a^2b - b + b^3)]/(a^2 + b^2)

Let the imaginary part equal to zero
a^2b - b + b^3 = 0
Divide by b
a^2 -1 + b^2 = 0
a^2 + b^2 = 1
As the modulus z is the square root of a^2 + b^2
Therefore l z l = 1

Sorry for the odd formatting.

 

[glittergal96]

Hey guys! It would really be great if you guys helped me out with a few questions that I can't do. If you're helping me, please show working out so I know how to get there!
Thanks!

The questions are :
1. If z is a complex number where neither Re(z) or Im(z) is zero and it is given that z+1/z is real, find [z] (absolute values bars)

Thanks guys!

If we let z = x + iy, we get:



As we are told that this quantity is real, it must have zero imaginary part. And since we also know that y is nonzero, we are left with

 

[wiliaamnguuyen]

Thanks a lot for your help! Yeah, I'm 50% finished with Complex Numbers. Just doing some practice questions, and this caught by eye. Thanks anyways!

 

[ledamn]

Since this is a kind of Complex Number marathon thread. Here's a few questions, for you geniuses!

 

[ledamn]

View attachment 30921

Since this is a kind of Complex Number marathon thread. Here's a few questions, for you geniuses!

 

[lpodtouch]

View attachment 30921

Since this is a kind of Complex Number marathon thread. Here's a few questions, for you geniuses!

For question two, is the answer z=2+i or z=2-3i

 

[ledamn]

How did you get to z = 2 + i , 2 - 3i?

 

[lpodtouch]

How did you get to z = 2 + i , 2 - 3i?

Is it correct?
If we were to tweak the equation, it would appear as:
x^2 + y^2 + 2i(x-iy) = 4i + 7

Separate into real and imaginary parts, then equate coefficients to receive the final answer (z=x+iy).

 

[ledamn]

Nice! It's absolutely correct!

 

[ledamn]

@Ipodtouch

For that question.
Isn't it
|a+ib|^2 + 2i(a-ib) = 4i + 7

(a+ib)^2 + 2ia +2b = 4i + 7
a^2 + 2iab - b^2 + 2ia + 2b = 4i + 7
2iab + 2ia + a^2 - b^2 + 2b = 4i + 7


Equate the real part and the imaginary part right?
(Re): a^2 - b^2 + 2b = 7
(Im): 2iab + 2ia = 4i
b + 1 = 2
b = 1
Sub b=1 into (Re)
a^2 - (1)^2 + 2(1) = 7
a^2 - 1 + 2 = 7
a^2 = 6
a = squareroot(6)

I don't think, I'm doing this right. What am I doing wrong?

 

[ledamn]

AHAHA! It's correct? The answers are: z= 2+i, 2-3i. Dammit, I can't get too it

 

[lpodtouch]

@Ipodtouch

For that question.
Isn't it
|a+ib|^2 + 2i(a-ib) = 4i + 7

(a+ib)^2 + 2ia +2b = 4i + 7
a^2 + 2iab - b^2 + 2ia + 2b = 4i + 7
2iab + 2ia + a^2 - b^2 + 2b = 4i + 7


Equate the real part and the imaginary part right?
(Re): a^2 - b^2 + 2b = 7
(Im): 2iab + 2ia = 4i
b + 1 = 2
b = 1
Sub b=1 into (Re)
a^2 - (1)^2 + 2(1) = 7
a^2 - 1 + 2 = 7
a^2 = 6
a = squareroot(6)

I don't think, I'm doing this right. What am I doing wrong?

The modulus of z is equal to the square root of a^2 + b^2 (a^2 + b^2)^(1/2)

If you were to square the modulus, it would end up as a^2 + b^2
Do you have the answer for the given question?

 

[lpodtouch]

AHAHA! It's correct? The answers are: z= 2+i, 2-3i. Dammit, I can't get too it

For the first question specified:

Let z= x+iy
l z l = square root of (x^2 + y^2)
l z l^2 = x^2 + y^2
x^2 + y^2 + 2i(x-iy) = 4i + 7
x^2 + 2y + y^2 + 2ix = 7 + 4i
Equating co-efficients:
x^2 + 2y + y^2 = 7 (1)
2ix=4i
x= 2
Sub x=2 into (1)
y^2 + 2y -3 = 0
(y+3)(y-1) = 0
y= -3, 1

Therefore, z = 2-3i or z= 2+i

 

[ledamn]

Answers are posted!

 

[ledamn]

Good answer @Ipodtouch! Anyone else wanna try the other questions?

 

[wiliaamnguuyen]

|z| = squareroot(1+ (squareroot(3)^2)
= squareroot(1+3)
= 2

 

[Tugga]

Hey guys! It would really be great if you guys helped me out with a few questions that I can't do. If you're helping me, please show working out so I know how to get there!
Thanks!

The questions are :
1. If z is a complex number where neither Re(z) or Im(z) is zero and it is given that z+1/z is real, find [z] (absolute values bars)

Thanks guys!

I think its easier to let z=rcis(theta) rather than a+ib. Using DMT gives r-(1/r)=0 which clearly means r=1 (since r>0)

 

[dunjaaa]

You can always verify your answer by doing sum and product of roots. Here is how you do 4(iii) . Parts (i) and (ii) are basic operations. To avoid confusion, let z(1)=x+iy and z(2)=a+ib and try working from there.