# complex numbers

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Kalabamboo

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#1

Hey guys,

so am struggling to understand this - can somebody help please? Thanks!!!

so am struggling to understand this - can somebody help please? Thanks!!!

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math42

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#2

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#2

(Original post by

Hey guys,

so am struggling to understand this - can somebody help please? Thanks!!!

**Kalabamboo**)Hey guys,

so am struggling to understand this - can somebody help please? Thanks!!!

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Kalabamboo

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#3

(Original post by

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.

**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.

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B_9710

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#4

(Original post by

Hey thanks! But what do you mean by they just have 0 imaginary part?

**Kalabamboo**)Hey thanks! But what do you mean by they just have 0 imaginary part?

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**Sinnoh**

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#5

(Original post by

Hey thanks! But what do you mean by they just have 0 imaginary part?

**Kalabamboo**)Hey thanks! But what do you mean by they just have 0 imaginary part?

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Kalabamboo

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#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.

I got confused with the bold bit - could you please help?

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?

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math42

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#7

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#7

(Original post by

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.

I got confused with the bold bit - could you please help?

**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?

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?

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#8

(Original post by

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

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

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?

**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?

- 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

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math42

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#9

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#9

(Original post by

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

**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

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Anonymouspsych

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#10

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#10

**Kalabamboo**)

Hey guys,

so am struggling to understand this - can somebody help please? Thanks!!!

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

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#11

(Original post by

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).

**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).

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?

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#12

(Original post by

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?

**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?

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#13

(Original post by

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).

**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).

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

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#14

**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).

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

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#15

(Original post by

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

**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

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#16

(Original post by

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.

**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.

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#17

(Original post by

Ah thank you so much - and you know i itself, is it like any imaginary number or is it just square root(-1)?

**Kalabamboo**)Ah thank you so much - and you know i itself, is it like any imaginary number or is it just square root(-1)?

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#18

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i = root(-1), yes.

**math42**)i = root(-1), yes.

you could just write complex number = x + square root(-1)y where x and y are real numbers ?

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#19

(Original post by

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 ?

**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 ?

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#20

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If you really wanted to, yes.

**math42**)If you really wanted to, yes.

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