# complex numbers

Announcements
#1
Hey guys,
so am struggling to understand this - can somebody help please? Thanks!!!
0
3 years ago
#2
(Original post by Kalabamboo)
Hey guys,
so am struggling to understand this - can somebody help please? Thanks!!!
To be honest, this answer seems insufficient. Real numbers are also complex; they just have 0 imaginary part. Clearly z + 2i can have 0 imaginary part if we choose z to have imaginary part -2i. But this is the only time when the equation can hold true, since |z| is real. Writing z = x + iy, see if you can derive a contradiction by plugging in |z| = root(x^2 + y^2), y = - 2.
2
#3
(Original post by math42)
To be honest, this answer seems insufficient. Real numbers are also complex; they just have 0 imaginary part. Clearly z + 2i can have 0 imaginary part if we choose z to have imaginary part -2i. But this is the only time when the equation can hold true, since |z| is real. Writing z = x + iy, see if you can derive a contradiction by plugging in |z| = root(x^2 + y^2), y = - 2.
Hey thanks! But what do you mean by they just have 0 imaginary part?
0
3 years ago
#4
(Original post by Kalabamboo)
Hey thanks! But what do you mean by they just have 0 imaginary part?
A real number can be viewed as a complex number with imaginary part 0. So a real number n can be viewed as n+0i.
1
3 years ago
#5
(Original post by Kalabamboo)
Hey thanks! But what do you mean by they just have 0 imaginary part?
All complex numbers can be written in the form where a and b are real integers. There's a real part and an imaginary part there. If b = 0, then there's no imaginary part.
0
#6
Thank you very much guys -this is all starting to make sense Ok so continuing on..from math42's post
"To be honest, this answer seems insufficient. Real numbers are also complex; they just have 0 imaginary part. Clearly z + 2i can have 0 imaginary part if we choose z to have imaginary part -2i. But this is the only time when the equation can hold true, since |z| is real. Writing z = x + iy, see if you can derive a contradiction by plugging in |z| = root(x^2 + y^2), y = - 2. "
I got confused with the bold bit - could you please help? 0
3 years ago
#7
(Original post by Kalabamboo)
Thank you very much guys -this is all starting to make sense Ok so continuing on..from math42's post
"To be honest, this answer seems insufficient. Real numbers are also complex; they just have 0 imaginary part. Clearly z + 2i can have 0 imaginary part if we choose z to have imaginary part -2i. But this is the only time when the equation can hold true, since |z| is real. Writing z = x + iy, see if you can derive a contradiction by plugging in |z| = root(x^2 + y^2), y = - 2. "
I got confused with the bold bit - could you please help? So we want to show that the equation |z| = z + 2i has no solutions, where z is a complex number.

To do this, we assume that it does have a solution, and see where this leads us. If it leads us to something which is clearly wrong/makes no sense, our initial assumption was wrong. This is a "proof by contradiction", which you may have seen.

Well, start by writing z = x + iy, where x,y are real (we know by definition that all complex numbers can be written this way). The equation for |z| is then |z| = root(x^2 + y^2) - it is the length of the line joining z to 0 in the complex plane.

So we have root(x^2 + y^2) = x + iy + 2i = x + i(y + 2).

The real and imaginary parts on each side have to match up. The left hand side is real, the right hand side has the imaginary bit i(y + 2). This must be 0 to match up with the left hand side, so y + 2 = 0, i.e. y = -2.

Then we get root(x^2 + y^2) = x. But y = -2, and (-2)^2 = 4, so root(x^2 + 4) = x. Can you see why this doesn't make sense?
0
#8
(Original post by math42)
So we want to show that the equation |z| = z + 2i has no solutions, where z is a complex number.

To do this, we assume that it does have a solution, and see where this leads us. If it leads us to something which is clearly wrong/makes no sense, our initial assumption was wrong. This is a "proof by contradiction", which you may have seen.

Well, start by writing z = x + iy, where x,y are real (we know by definition that all complex numbers can be written this way). The equation for |z| is then |z| = root(x^2 + y^2) - it is the length of the line joining z to 0 in the complex plane.

So we have root(x^2 + y^2) = x + iy + 2i = x + i(y + 2).

The real and imaginary parts on each side have to match up. The left hand side is real, the right hand side has the imaginary bit i(y + 2). This must be 0 to match up with the left hand side, so y + 2 = 0, i.e. y = -2.

Then we get root(x^2 + y^2) = x. But y = -2, and (-2)^2 = 4, so root(x^2 + 4) = x. Can you see why this doesn't make sense?
Hey thanks a lot! But you know where you said |z| = root(x^2 + y^2)
- isn't it |z| = root(1^2 + i^2) because z = x + iy ?
Btw im thinking about the i + j + z notation... but in this case i is a number
0
3 years ago
#9
(Original post by Kalabamboo)
Hey thanks a lot! But you know where you said |z| = root(x^2 + y^2)
- isn't it |z| = root(1^2 + i^2) because z = x + iy ?
Btw im thinking about the i + j + z notation... but in this case i is a number
No. It seems you're getting mixed up with general vectors. |z| is the length of the line from 0 to z. You can split this into horizontal and vertical components x and y (remember, the vertical axis is the imaginary axis, so you don't need to worry about including i in these calculations). Then just apply pythagoras to get root(x^2 + y^2).

0
3 years ago
#10
(Original post by Kalabamboo)
Hey guys,
so am struggling to understand this - can somebody help please? Thanks!!!
But z can be a complex number that has an imaginary part equal to -2i so that overall z+2i is real so your reasoning is not sufficient (in order to ensure the RHS is real you need the imaginary part of z to cancel out with the +2i which would obviously be -2i)

If we assume z = a-2i where a is the real part, then RHS = z+2i = a-2i+2i=a which is real
LHS = abs(z) = abs(a-2i) = sqrt (a^2+4)

So you are saying sqrt (a^2+4) = a
-----> a^2+4 = a^2
-----> 4=0 which is clearly not true so hence no solution

Note that
0
#11
(Original post by math42)
No. It seems you're getting mixed up with general vectors. |z| is the length of the line from 0 to z. You can split this into horizontal and vertical components x and y (remember, the vertical axis is the imaginary axis, so you don't need to worry about including i in these calculations). Then just apply pythagoras to get root(x^2 + y^2).

Yes you're right! I am confusing it with vectors.
But then what form is z = x + iy written in? Is this something I would've done in A level maths cause it's not in vector form but it is so similar to it and i can't really compare it to anything else.
Could you pls kindly lemme know your thoughts?
0
3 years ago
#12
(Original post by Kalabamboo)
Yes you're right! I am confusing it with vectors.
But then what form is z = x + iy written in? Is this something I would've done in A level maths cause it's not in vector form but it is so similar to it and i can't really compare it to anything else.
Could you pls kindly lemme know your thoughts?
Fundamentally, x + iy is just the standard form for complex numbers, real part plus imaginary part. But for the purposes of calculation you can liken it to the standard vectors you know if you like. In 2 dimensions you have the unit vectors i and j right? Well, here, your unit vectors would instead be, correspondingly, 1 and i, if that makes sense (don't confuse the two different "i"s; they aren't related to each other). So x + iy in the complex plane is like the vector xi + yj in the standard plane, which I assume you know has length root(x^2 + y^2).
0
#13
(Original post by math42)
Fundamentally, x + iy is just the standard form for complex numbers, real part plus imaginary part. But for the purposes of calculation you can liken it to the standard vectors you know if you like. In 2 dimensions you have the unit vectors i and j right? Well, here, your unit vectors would instead be, correspondingly, 1 and i, if that makes sense (don't confuse the two different "i"s; they aren't related to each other). So x + iy in the complex plane is like the vector xi + yj in the standard plane, which I assume you know has length root(x^2 + y^2).
ah thanks so much!! so sorry my laptop decided to update at the wrong time
so I get what you did there - was wondering so you know how you said
"x + iy is just the standard form for complex numbers, real part plus imaginary part. " so the real part is always in the x direction and the imaginary part is always in the y direction? Also what's the point in having iy - why can't it just be i or y?

Thanks so much for answering my silly questions so far It's just that I am quite new with complex numbers and the way they are written
0
#14
(Original post by math42)
Fundamentally, x + iy is just the standard form for complex numbers, real part plus imaginary part. But for the purposes of calculation you can liken it to the standard vectors you know if you like. In 2 dimensions you have the unit vectors i and j right? Well, here, your unit vectors would instead be, correspondingly, 1 and i, if that makes sense (don't confuse the two different "i"s; they aren't related to each other). So x + iy in the complex plane is like the vector xi + yj in the standard plane, which I assume you know has length root(x^2 + y^2).
ah thanks so much!! so sorry my laptop decided to update at the wrong time
so I get what you did there - was wondering so you know how you said
"x + iy is just the standard form for complex numbers, real part plus imaginary part. " so the real part is always in the x direction and the imaginary part is always in the y direction? Also what's the point in having iy - why can't it just be i or y?

Thanks so much for answering my silly questions so far
It's just that I am quite new with complex numbers and the way they are written
0
3 years ago
#15
(Original post by Kalabamboo)
ah thanks so much!! so sorry my laptop decided to update at the wrong time
so I get what you did there - was wondering so you know how you said
"x + iy is just the standard form for complex numbers, real part plus imaginary part. " so the real part is always in the x direction and the imaginary part is always in the y direction? Also what's the point in having iy - why can't it just be i or y?

Thanks so much for answering my silly questions so far
It's just that I am quite new with complex numbers and the way they are written
Yes, the real part is always in the x direction and the imaginary part is always in the y direction. Well, the "i" tells you that it's the imaginary component, the y tells you how big it is. If it was just x + y it would just be a real number. Complex numbers are precisely numbers of the form x + iy for x,y real.
0
#16
(Original post by math42)
Yes, the real part is always in the x direction and the imaginary part is always in the y direction. Well, the "i" tells you that it's the imaginary component, the y tells you how big it is. If it was just x + y it would just be a real number. Complex numbers are precisely numbers of the form x + iy for x,y real.
Ah thank you so much - and you know i itself, is it like any imaginary number or is it just square root(-1)?
0
3 years ago
#17
(Original post by Kalabamboo)
Ah thank you so much - and you know i itself, is it like any imaginary number or is it just square root(-1)?
i = root(-1), yes.
0
#18
(Original post by math42)
i = root(-1), yes.
oooh so instead of writing complex number = x + iy
you could just write complex number = x + square root(-1)y where x and y are real numbers ?
0
3 years ago
#19
(Original post by Kalabamboo)
oooh so instead of writing complex number = x + iy
you could just write complex number = x + square root(-1)y where x and y are real numbers ?
If you really wanted to, yes.
0
#20
(Original post by math42)
If you really wanted to, yes.
Ahh thank you so much!
1
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How did your Edexcel GCSE Maths Paper 1 go?

Loved the paper - Feeling positive (30)
18.99%
The paper was reasonable (67)
42.41%
Not feeling great about that exam... (28)
17.72%
It was TERRIBLE (33)
20.89%