Complex number calculator necessary for maths ext 2? (1 Viewer)

[Beca-W]

Hey everyone,

I'm just wondering whether it would be advantageous to get a complex number calculator for 4U maths?

Thanks!

 

[DatAtarLyfe]

No. If you need a cal to do basic complex numbers you shouldn't be doing 4u

 

[pikachu975]

Hey everyone,

I'm just wondering whether it would be advantageous to get a complex number calculator for 4U maths?

Thanks!

It's not really necessary but I guess you can check your complex number arithmetic or mod-arg conversions.

 

[leehuan]

No. If you need a cal to do basic complex numbers you shouldn't be doing 4u

Maybe if you're relying on it to do them. But what's the harm in getting one to check your answer so that you can guarantee those marks?

 

[Sp3ctre]

Hey everyone,

I'm just wondering whether it would be advantageous to get a complex number calculator for 4U maths?

Thanks!

It might be advantageous for you to easily get those complex number calculations done without spending too much time on the algebra side, but in the long-run it could potentially have a negative outcome of your algebraic skills. I would just stay on the same level as the other students in your class (assuming they don't all have complex number calculators) just to get used to the time it takes to do some complex number calculations, since time management will be very important during an exam.

 

[D94]

Maybe if you're relying on it to do them. But what's the harm in getting one to check your answer so that you can guarantee those marks?

Because a calculator is only as good as your input. GIGO

There are other tests to determine the correctness of your solution. It's more important for a student to learn these than to use a calculator since you won't have access to those calculators in an exam.

 

[pikachu975]

Because a calculator is only as good as your input. GIGO

There are other tests to determine the correctness of your solution. It's more important for a student to learn these than to use a calculator since you won't have access to those calculators in an exam.

The fx 100 AU plus is allowed in exams and it's a complex number calculator though.

 

[D94]

The fx 100 AU plus is allowed in exams and it's a complex number calculator though.

I meant in general on the reliance of calculators.

 

[Silly Sausage]

Quite handy in certain situations only e.g. finding the complex roots of a quadratic equation, won't give you much of an advantage but still comes useful sometimes.

 

[lazzzy]

Quite handy in certain situations only e.g. finding the complex roots of a quadratic equation, won't give you much of an advantage but still comes useful sometimes.

Curious as to how you would do this? (on the fx 100 au plus)?

 

[pikachu975]

I have the complex number calculator it's not as useful as it seems

 

[marxman]

I used mine quite a bit and in reality there is no serious disadvantage to owning one. It can help you get 1 mark questions in a matter of seconds and also check your arithmetic on any yucky computations.

From a competitive perspective, if it's allowed to be used then you can assume other people will use it, and these people will have the advantage of being able to do calculations faster than you. It's unnecessary to put yourself at this kind of disadvantage (unless you are an absolute complex arithmetic genius).

 

[dan964]

For those who have the fx82AU or fx100au, you can use the Pol() and Rec() functions (learn how to use them), to convert a complex number from x+iy form
to polar form. Just simply convert the number from

So write Pol(x,y) gives you the modulus and the angle which are stored to X and Y variables on your calculator respectively. (Modulus is stored as the X variable and as the Ans key, you can then square it to find what it is the squareroot of; and for the Angle typing in Y and then pressing the relevant key can convert it into degrees and minutes (if calculator is set to such).

To go the other way put Rec(|z|, argz), gives you x and y, which are stored under X and Y variables.


For complex roots, remember that if you are solving a quadratic equation with real coefficients, both the root and its conjugate are roots. The easier way to find quadratic roots is to simply use the quadratic formula.

Firstly compute (the discriminate) if this is negative then you have two complex roots...

So then if that is the case, if you can complete the square to the form , then the complex number a+ib and a-ib are roots. (This is derivated from

 

[pikachu975]

For those who have the fx82AU or fx100au, you can use the Pol() and Rec() functions (learn how to use them), to convert a complex number from x+iy form
to polar form. Just simply convert the number from

So write Pol(x,y) gives you the modulus and the angle which are stored to X and Y variables on your calculator respectively. (Modulus is stored as the X variable and as the Ans key, you can then square it to find what it is the squareroot of; and for the Angle typing in Y and then pressing the relevant key can convert it into degrees and minutes (if calculator is set to such).

To go the other way put Rec(|z|, argz), gives you x and y, which are stored under X and Y variables.


For complex roots, remember that if you are solving a quadratic equation with real coefficients, both the root and its conjugate are roots. The easier way to find quadratic roots is to simply use the quadratic formula.

If you can complete the square to the form , then the complex number a+ib and a-ib are roots. (This is derivated from

Just a question unrelated, but about square roots:

When square rooting a complex number e.g. 8 + 6i
So we use the form a^2 +2abi - b^2
sqrt [3^2 - 1^2 + 2*3*i] = +/- (3+i)

Is this quick method allowed for like 2-3 markers for square rooting or is it only for multiple choice? Is it allowed or do we have to do the long method:
a^2 + 2abi - b^2 = 8 + 6i
Equating real and imaginary parts....

Thanks!

 

[InteGrand]

Just a question unrelated, but about square roots:

When square rooting a complex number e.g. 8 + 6i
So we use the form a^2 +2abi - b^2
sqrt [3^2 - 1^2 + 2*3*i] = +/- (3+i)

Is this quick method allowed for like 2-3 markers for square rooting or is it only for multiple choice? Is it allowed or do we have to do the long method:
a^2 + 2abi - b^2 = 8 + 6i
Equating real and imaginary parts....

Thanks!

In other words, inspection. Should be fine as long as it's clear what you did.

 

[dan964]

Just a question unrelated, but about square roots:

When square rooting a complex number e.g. 8 + 6i
So we use the form a^2 +2abi - b^2
sqrt [3^2 - 1^2 + 2*3*i] = +/- (3+i)

Is this quick method allowed for like 2-3 markers for square rooting or is it only for multiple choice? Is it allowed or do we have to do the long method:
a^2 + 2abi - b^2 = 8 + 6i
Equating real and imaginary parts....

Thanks!

Inspection works with integers/simple fractions.