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    Could someone spare a moment to explain how they are done. I am learning on my own and don't understand part 3

    3b) Write in the form a +ib:

    6(cos3pi/4 + i sin 3pi/4)


    Thanks

    Also if anyone has the book, question 1b on Exercise 1F too...

    Thanks
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    Evaluate \cos \frac{3\pi}{4} and \sin \frac{3\pi}{4} and expand the brackets... what don't you understand about it?

    I don't have this book, and it's likely that the majority of people here don't, so if you want help on questions it's best to post them.
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    (Original post by nuodai)
    Evaluate \cos \frac{3\pi}{4} and \sin \frac{3\pi}{4} and expand the brackets... what don't you understand about it?

    I don't have this book, and it's likely that the majority of people here don't, so if you want help on questions it's best to post them.
    Well i know how to go from a +ib to r(cos? +i sin?) etc but don't understand how to do the reverse.

    As for 1B: Finding the argument for 3i. I can do it when it is 2+2i for example but not when there is only x or y.

    Thanks
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    (Original post by Albe-Ube)
    Well i know how to go from a +ib to r(cos? +i sin?) etc but don't understand how to do the reverse.

    As for 1B: Finding the argument for 3i. I can do it when it is 2+2i for example but not when there is only x or y.

    Thanks
    Just expand the brackets and evaluate the trig terms, as nuodai said.

    For the second bit, draw it on an argand diagram to help. What's the anticlockwise angle between the real axis and 3i?
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    (Original post by Scipio90)
    Just expand the brackets and evaluate the trig terms, as nuodai said.

    For the second bit, draw it on an argand diagram to help. What's the anticlockwise angle between the real axis and 3i?
    Well i think of 2 + 4i as co-ordinates 2,4.

    So for 3i, i thought it was 0,3. But that means it lies on the axis, instead of being in one of the four quadrants?
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    (Original post by Albe-Ube)
    Well i think of 2 + 4i as co-ordinates 2,4.

    So for 3i, i thought it was 0,3. But that means it lies on the axis, instead of being in one of the four quadrants?
    Yes. That is not a problem. What's the angle?
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    (Original post by rnd)
    Yes. That is not a problem. What's the angle?
    that's what i don't understand!

    90 degrees?

    That means tan of the angle = pi/2?
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    (Original post by Albe-Ube)
    Well i think of 2 + 4i as co-ordinates 2,4.

    So for 3i, i thought it was 0,3. But that means it lies on the axis, instead of being in one of the four quadrants?
    Don't worry about quadrants; all you have to worry about is the angle that a point makes with the positive x-axis. You can use either -\pi < \theta \le \pi (which means the 3rd and 4th quadrants take negative values) or 0 \le \theta < 2\pi (which means that the 3rd and 4th quadrant take positive values greater than 180^{\circ}.

    So, if k \ge 0, if a point lied on the positive x-axis, that is k + 0i, it would have argument 0. If it were on the positive y-axis, that is 0 + ki, it would have argument \frac{\pi}{2}. If it were on the negative x-axis, that is -k + 0i it would have argument \pi. If it were on the negative y-axis, that is 0 - ki it would have argument \frac{3\pi}{2} or -\frac{\pi}{2} depending on convention (you can check that both give the same result).

    A quick rule is that if you have a complex number x + iy, then its modulus is r = \sqrt{x^2 + y^2} and its argument is \theta = \tan^{-1} \left( \dfrac{y}{x} \right), such that x + iy = r(\cos \theta + i\sin \theta) = re^{i\theta} (ignore the re^{i\theta} if you haven't or don't need to cover that).

    This general result is proved quite easily. If your complex number z = x + iy is represented by point Z(x, y) you drew a line from the origin to the point Z and from there vertically down to the point (x, 0), you'd have a right-angled triangle. The argument is the angle ZOP (which can be found using trig), and the modulus is the length of the hypotenuse, which by Pythagoras's Theorem must be \sqrt{x^2 + y^2}.
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    Thanks alot! Understand it now
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    (Original post by Albe-Ube)
    that's what i don't understand!

    90 degrees?

    That means tan of the angle = pi/2?
    No.
    The tan of the angle is not is pi/2.
    The angle is pi/2.

    Sorry Nuodai, if that's already there in your post.
    I thought it needed to be put in a short, simple post.
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    Argh I am struggling. I still don't get question 3.

    Can you just do this one for me so then I'll be able to do the rest.

    Write in the form a + ib,

    6( cos 3pi/4 + i sin 3pi/4)
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    (Original post by Albe-Ube)
    Argh I am struggling. I still don't get question 3.

    Can you just do this one for me so then I'll be able to do the rest.

    Write in the form a + ib,

    6( cos 3pi/4 + i sin 3pi/4)
    a=6cos(3pi/4)
    b=6sin(3pi/4)
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    (Original post by rnd)
    a=6cos(3pi/4)
    b=6sin(3pi/4)
    I mean could you just do me a big favour and answer the question. The answer is -3root2 + 3root2i if that helps.

    Thanks
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    (Original post by Albe-Ube)
    I mean could you just do me a big favour and answer the question. The answer is -3root2 + 3root2i if that helps.

    Thanks
    :confused:

    You have the answer. There are no steps missed out apart from working out 6cos(3pi/4) etc..

    Could you be clearer about what's bothering you?
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    (Original post by rnd)
    :confused:

    You have the answer. There are no steps missed out apart from working out 6cos(3pi/4) etc..

    Could you be clearer about what's bothering you?
    haha i've got it now Thanks alot.

    Moved on and powering through 1G now.

    If i get stuck further on in the book, maybe I'll ask fr more help!

    Thanks again
 
 
 
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