# Complex Number - cos z

Thread starter 7 years ago
#1
This is an IB Higher Level question, it is talking about cos z and how it is equal to 2. This begs two questions: How can the z be a complex number - shouldn't it be an angle (what is the argument therefore, an angle of an angle???). And how can it be 2, when it is usually between -1 and 1?

Full question attached...

Any hint would be appreciated
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7 years ago
#2
The question is imply referring to cos(z) as a function of a number, don't think of z as an angle. Think of it as an infinite MacLaurin series as opposed to a function of an angle to do with circles and triangles. When complex numbers get involved, cos(z) doesn't have to be between -1 and 1
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7 years ago
#3
(Original post by Gmart)
This is an IB Higher Level question, it is talking about cos z and how it is equal to 2. This begs two questions: How can the z be a complex number - shouldn't it be an angle (what is the argument therefore, an angle of an angle???). And how can it be 2, when it is usually between -1 and 1?

Full question attached...

Any hint would be appreciated
Do you recognise this: ?
If so, we can do this:

Then adding them together, we get: .
So as long as you believe the first line is true, then the rest of it follows from that, so cos(z) where z is complex (not an angle) is perfectly fine to write.

But to answer the question, ignore the fact it says cos(z) and try to solve: .
I find it helps to write .
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Thread starter 7 years ago
#4
OK, i've solved it (yay!!) but i am still a bit stuck on the two questions... i will think about it
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7 years ago
#5
(Original post by Gmart)
OK, i've solved it (yay!!) but i am still a bit stuck on the two questions... i will think about it
In the expression cos z, where z can be complex, z should no longer be thought of as angle - just a value you substitute into the function.

If z is allowed to be complex, cos z can take on any (finite) value. This takes a bit of getting used to, but if you use the definition of cos z in terms of exponentials you can form an equation involving which allows you to solve cos z = w for w > 1.
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Thread starter 7 years ago
#6
(Original post by davros)
In the expression cos z, where z can be complex, z should no longer be thought of as angle - just a value you substitute into the function.

If z is allowed to be complex, cos z can take on any (finite) value. This takes a bit of getting used to, but if you use the definition of cos z in terms of exponentials you can form an equation involving which allows you to solve cos z = w for w > 1.
I can see that that must be the case, but I am still somewhat confused by it. Is it still cosine, in that it is the adjacent over hypoteneuse of a rt triangle? Or is it some other function which happens to be called cosine (why). How is it useful to have it with w>1? Why do we need it? When? are all the trig ratios also able to have a complex number instead of an angle? how do they relate to each other? Do they relate to each other? If i square this complex number and add the square of the same sine complex number, will it equal 1? If I graph it, what will happen (my graphic calculator is not keen, talks about it being non-real and therefore not allowed...)

Many questions, but if i have to accept that these answers are at a higher level, then where? Which?
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7 years ago
#7
I need to brush up on complex numbers but can you represent z as a complex number z=x+-iy and using the compound angle formula solve the equation by equating real and imaginary parts?
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7 years ago
#8
(Original post by Gmart)
I can see that that must be the case, but I am still somewhat confused by it. Is it still cosine, in that it is the adjacent over hypoteneuse of a rt triangle? Or is it some other function which happens to be called cosine (why). How is it useful to have it with w>1? Why do we need it? When? are all the trig ratios also able to have a complex number instead of an angle? how do they relate to each other? Do they relate to each other? If i square this complex number and add the square of the same sine complex number, will it equal 1? If I graph it, what will happen (my graphic calculator is not keen, talks about it being non-real and therefore not allowed...)

Many questions, but if i have to accept that these answers are at a higher level, then where? Which?
Many questions

The school way (and historical way) of teaching sin and cos is:

first define them in terms of angles in a right-angled triangle, so they only make sense for angles between 0 and 90 degrees;

THEN extend the definition using a point going round the unit circle, so we can make sense of sin, cos (and tan) for real angles, positive and negative, of any size by interpreting cos and sin as coordinates of a point as it goes round

THEN extend the definition again by introducing Euler's formula for complex exponentials.

This is all very piecemeal and disjointed, but it's the only realistic way to do things with a school level of mathematics.

At university level, we define sin and cos 'analytically', using something called a power series - look this up if you haven't seen it, it's basically a more general version of the infinite series you've probably seen when looking at GPs or the binomial expansion. Power series give us all sorts of wonderful properties like continuity and differentiability, but they don't really know anything about 'angles' so for example the number pi has to be defined analytically - e.g. we can say that pi/2 is the smallest positive solution of cos z = 0.

Once we've sorted this out, we can show that our power series for sin and cos satisfy all the desirable properties we want from them - e.g. for all z, real or complex; if z is real then |sin z| <= 1 and |cos z| <= 1 but if z is complex we can make sin and cos as big as we want (and they can be complex too!).

Graphical calculators are very powerful now, but I'm not sure if they can handle sin/cos for complex numbers. If you think about trying to visualize it, you have 2 coordinates x and y in x +yi to feed into the function, and the output can also be complex u + vi with 2 more coordinates! So it's not an easy thing to visualize

Many functions that you have met already just in terms of real numbers can be extended to deal with complex numbers too - in some ways it's more natural to do this and it gives you lots of powerful results.
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7 years ago
#9
(Original post by Davelittle)
I need to brush up on complex numbers but can you represent z as a complex number z=x+-iy and using the compound angle formula solve the equation by equating real and imaginary parts?
You can do it this way, but it's easier to see it as a quadratic and solve that way.
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7 years ago
#10
(Original post by Gmart)
This is an IB Higher Level question, it is talking about cos z and how it is equal to 2. This begs two questions: How can the z be a complex number - shouldn't it be an angle (what is the argument therefore, an angle of an angle???). And how can it be 2, when it is usually between -1 and 1?

Full question attached...

Any hint would be appreciated
look at Q154 in this link

In this booklet I am solving sin z = 2
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