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Tangents parallel to the initial line: consider
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Tangents perpendicular to the initial line: consider
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Area under a curve (in formula book).
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Rearrange the transformation to give something in terms of w.
•
Take the modulus of both sides. If you have a fraction (which you commonly will), you can take the modulus of the top and the bottom separately to multiply both sides by the denominator. If you have a negative sign in front of the fraction, you can drop that here as we're taking the absolute value.
•
Using , you can make your substitution so that you have an equation entirely in w (or the plane you're transforming to).
•
Now, . Make this substitution and group the real and imaginary terms in the moduli. You may find it easy to take a common factor of i if there is one.
•
You can then treat both sides as a complex number and take the modulus of them (Pythagoras). This means: Most likely, you will end up with a quadratic in terms of u or v if there's been multiple parts to a real or imaginary part.
•
Square both sides, and rearrange until the required form is achieved. I love this step cos loads of terms cancel down :-)
•
Complete the square if it needs to be a circle.
•
Rearrange the transformation to give something in terms of w.
•
Take the modulus of both sides. If you have a fraction (which you commonly will), you can take the modulus of the top and the bottom separately to multiply both sides by the denominator. If you have a negative sign in front of the fraction, you can drop that here as we're taking the absolute value.
•
Using , you can make your substitution so that you have an equation entirely in w (or the plane you're transforming to).
•
Now, . Make this substitution and group the real and imaginary terms in the moduli. You may find it easy to take a common factor of i if there is one.
•
You can then treat both sides as a complex number and take the modulus of them (Pythagoras). This means: Most likely, you will end up with a quadratic in terms of u or v if there's been multiple parts to a real or imaginary part.
•
Square both sides, and rearrange until the required form is achieved. I love this step cos loads of terms cancel down :-)
•
Complete the square if it needs to be a circle.
•
Rearrange the transformation to give something in terms of w.
•
Take the modulus of both sides. If you have a fraction (which you commonly will), you can take the modulus of the top and the bottom separately to multiply both sides by the denominator. If you have a negative sign in front of the fraction, you can drop that here as we're taking the absolute value.
•
Using , you can make your substitution so that you have an equation entirely in w (or the plane you're transforming to).
•
Now, . Make this substitution and group the real and imaginary terms in the moduli. You may find it easy to take a common factor of i if there is one.
•
You can then treat both sides as a complex number and take the modulus of them (Pythagoras). This means: Most likely, you will end up with a quadratic in terms of u or v if there's been multiple parts to a real or imaginary part.
•
Square both sides, and rearrange until the required form is achieved. I love this step cos loads of terms cancel down :-)
•
Complete the square if it needs to be a circle.
•
As you said, you would subtract the two arguments if you had a division inside one argument.
•
Let one of these arguments equal some angle , and let the other equal some other angle .
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The difference in these two angles is the main angle from the original problem,
•
gives a positive angle. Therefore the angle must always be greater than .
•
Since is less than 90 degrees ( radians), the locus of z as P varies is on the major arc of the circle
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That is to say that if is acute, the locus is on the major arc of the circle. If is obtuse, the locus lies on the minor arc of the circle.
•
Now, since is always greater than , the lines do not intersect where you'd normally expect. You have to 'go back' on yourself to the point where they intersect in the negative axis.
•
You can then draw the locus as shown on the diagram.
•
As you said, you would subtract the two arguments if you had a division inside one argument.
•
Let one of these arguments equal some angle , and let the other equal some other angle .
•
The difference in these two angles is the main angle from the original problem,
•
gives a positive angle. Therefore the angle must always be greater than .
•
Since is less than 90 degrees ( radians), the locus of z as P varies is on the major arc of the circle
•
That is to say that if is acute, the locus is on the major arc of the circle. If is obtuse, the locus lies on the minor arc of the circle.
•
Now, since is always greater than , the lines do not intersect where you'd normally expect. You have to 'go back' on yourself to the point where they intersect in the negative axis.
•
You can then draw the locus as shown on the diagram.
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Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71