Who am I? Hi! My name is Tom and I am currently on a gap year with unconditional offers from Bristol and Durham to study Maths next year. Ever since I can remember maths has been my favourite subject and I loved the problem solving that comes with it. When in my final years of school I helped to run a maths mentoring scheme where anybody could come along to try some difficult maths or if they needed help with something from the classroom. I’m not just a bit geeky, I do love playing sports too! I’m a really keen hockey player and have played divisional level for the West of England and currently play for Cannock. I have helped coach hockey for lots of different age groups and love the enthusiasm people bring to sessions. Our sessions The best thing about maths is the sense of achievement in finishing a challenging problem, and I want that to be the main goal of our sessions. The best thing about these sessions will be that you decide the topics that you want to cover. We can have a quick chat about what you understand about the topic already and then move from the basics upwards. This way there is no confusion of terms and techniques between me and any of your teachers at school! I understand that maths is not everybody’s cup of tea, therefore I will try to make these sessions as friendly and fun as I can so that you don’t see them as a chore. Being able to tackle something that you find tricky is far more rewarding than something you find easy!Who am I? Hi! My name is Tom and I am currently on a gap year with unconditional offers from Bristol and Durham to study Maths next year. Ever since I can remember maths has been my favourite subject and I loved the problem solving that comes with it. When in my final years of school I helped to run a maths mentoring scheme where anybody could come along to try some difficult maths or if they needed help with something from the classroom. I’m not just a bit geeky, I do love playing sports too! I’m a really keen hockey player and have played divisional level for the West of England and currently play for Cannock. I have helped coach hockey for lots of different age groups and love the enthusiasm people bring to sessions. Our sessions The best thing about maths is the sense of achievement in finishing a challenging problem, and I want that to be the main goal of our sessions. The best thing about these sessions will be that you decide the topics that you want to cover. We can have a quick chat about what you understand about the topic already and then move from the basics upwards. This way there is no confusion of terms and techniques between me and any of your teachers at school! I understand that maths is not everybody’s cup of tea, therefore I will try to make these sessions as friendly and fun as I can so that you don’t see them as a chore. Being able to tackle something that you find tricky is far more rewarding than something you find easy!

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In order to understand this question we must define what modulus-argument form is. The modulus of a complex number is its distance from the origin (0,0) on the Argand Diagram. It is written as |z|.
The argument of a complex number is the angle subtended anticlockwise between the x axis and a line drawn from the origin to the complex number. This is written where the angle is between -π and π radians (where the angle is below the x axis, it is written as a negative because the angle is measured clockwise around the origin). It is written as arg(z).
If you are unaware of what the Argand diagram is, don’t worry, here is a brief explanation!
The Argand diagram is a really useful visual aid for the use of complex numbers where they are split into their real and imaginary components. The real part is represented by a value on the x axis and the imaginary part is represented by a value on the y axis. This allows us to visualise the “size” of complex numbers, in other words the modulus.
The number can be easily shown on the Argand diagram, with the x (real part) value of 1, and the y (imaginary part) value of √3.
To find the modulus of this number which we will now refer to as z, we must effectively find the length of the line drawn from the origin to z. By creating a right angled triangle with the modulus as the hypotenuse, we can see that the other two lengths are 1 and √3.
Pythagoras proved that a^2 +b^2=c^2 which we can use to find the length of the hypotenuse: the modulus. So the |z|^2=(1)^2 +(√3)^2
|z|^2=1+3=4
|z|=2
So we have found the modulus to be 2.
We have defined that arg(z) is the angle between the x axis and the line from the origin to z.
Using trigonometric properties of the right angled triangle drawn we can use the tangent function to find the argument.
We know that tan(x)=Opposite/Adjacent.
The opposite in this case is √3 and the adjacent is 1. Therefore tan(x)=√3/1=√3
Using the inverse tangent function we find that arg(z)=arctan(√3)= π/3 radians.
And there is our final answer, that |z|=2 and arg(z)= π/3.
Extension: What is the modulus-argument form of z= -1-2i
Answer |z|=√5
arg(z)=-2π/3In order to understand this question we must define what modulus-argument form is. The modulus of a complex number is its distance from the origin (0,0) on the Argand Diagram. It is written as |z|.
The argument of a complex number is the angle subtended anticlockwise between the x axis and a line drawn from the origin to the complex number. This is written where the angle is between -π and π radians (where the angle is below the x axis, it is written as a negative because the angle is measured clockwise around the origin). It is written as arg(z).
If you are unaware of what the Argand diagram is, don’t worry, here is a brief explanation!
The Argand diagram is a really useful visual aid for the use of complex numbers where they are split into their real and imaginary components. The real part is represented by a value on the x axis and the imaginary part is represented by a value on the y axis. This allows us to visualise the “size” of complex numbers, in other words the modulus.
The number can be easily shown on the Argand diagram, with the x (real part) value of 1, and the y (imaginary part) value of √3.
To find the modulus of this number which we will now refer to as z, we must effectively find the length of the line drawn from the origin to z. By creating a right angled triangle with the modulus as the hypotenuse, we can see that the other two lengths are 1 and √3.
Pythagoras proved that a^2 +b^2=c^2 which we can use to find the length of the hypotenuse: the modulus. So the |z|^2=(1)^2 +(√3)^2
|z|^2=1+3=4
|z|=2
So we have found the modulus to be 2.
We have defined that arg(z) is the angle between the x axis and the line from the origin to z.
Using trigonometric properties of the right angled triangle drawn we can use the tangent function to find the argument.
We know that tan(x)=Opposite/Adjacent.
The opposite in this case is √3 and the adjacent is 1. Therefore tan(x)=√3/1=√3
Using the inverse tangent function we find that arg(z)=arctan(√3)= π/3 radians.
And there is our final answer, that |z|=2 and arg(z)= π/3.
Extension: What is the modulus-argument form of z= -1-2i
Answer |z|=√5
arg(z)=-2π/3