- #1
Alshia
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1. Question
Note: arg ( ) refers to the argument of the complex number, i.e. the angle. | | refers to the modulus, the length.
Since I don't know how to show an Argand diagram here, I will describe my solution algebraically.1. (a)
arg(z - 3) = -3∏/4, arg(z + 3) = -∏/2
1. (b)
arg(z - 4i) = ∏, |z+6| = 5
2. Attempt
1. (a)
arg(z+3) creates an angle of 90° clockwise, so z must be complex number of the form z = -a-bi (via the parallelogram rule).
Since z + 3 lies on the y-axis, -a = -3
so z = -3 +(-b)i
Let p be a complex number where p = 0+ci, c<0. P is passed through by the vector (z-3).
arg(p-3) = -(∏-3∏/4) = -∏/4
tan (-∏/4) = p/3
p = -3
Now, since z is a complex number such that z = -3 - bi, p must be the midpoint of of (z-3).
So z = -3+(-3*2)i
∴ z = -3 - 6i1. (b)
z + 6 creates a straight line, so
arg(z + 6) = 0
z+6 = 5(cos 0 + i sin 0) = 5
z = -1
Correct answer: -3 + 4i, -9 + 4i3. My Issues
Is the working correct? My answer for 1.(a) is the same as the book, but I'm not sure whether the solution is "formal" enough.
For 1.(b), I got stuck and don't know how to solve the rest of it.
I am having problems geometrically and algebraically interpreting pairs of equations like arg(z+3)=a and arg(z-6)=b, where I have to find the points satisfying the solutions.
Does anyone know where I can find videos for this type of question? Most of the videos I've found merely expounds on the very basics, where z is given.
Thanks!
Note: arg ( ) refers to the argument of the complex number, i.e. the angle. | | refers to the modulus, the length.
Since I don't know how to show an Argand diagram here, I will describe my solution algebraically.1. (a)
arg(z - 3) = -3∏/4, arg(z + 3) = -∏/2
1. (b)
arg(z - 4i) = ∏, |z+6| = 5
2. Attempt
1. (a)
arg(z+3) creates an angle of 90° clockwise, so z must be complex number of the form z = -a-bi (via the parallelogram rule).
Since z + 3 lies on the y-axis, -a = -3
so z = -3 +(-b)i
Let p be a complex number where p = 0+ci, c<0. P is passed through by the vector (z-3).
arg(p-3) = -(∏-3∏/4) = -∏/4
tan (-∏/4) = p/3
p = -3
Now, since z is a complex number such that z = -3 - bi, p must be the midpoint of of (z-3).
So z = -3+(-3*2)i
∴ z = -3 - 6i1. (b)
z + 6 creates a straight line, so
arg(z + 6) = 0
z+6 = 5(cos 0 + i sin 0) = 5
z = -1
Correct answer: -3 + 4i, -9 + 4i3. My Issues
Is the working correct? My answer for 1.(a) is the same as the book, but I'm not sure whether the solution is "formal" enough.
For 1.(b), I got stuck and don't know how to solve the rest of it.
I am having problems geometrically and algebraically interpreting pairs of equations like arg(z+3)=a and arg(z-6)=b, where I have to find the points satisfying the solutions.
Does anyone know where I can find videos for this type of question? Most of the videos I've found merely expounds on the very basics, where z is given.
Thanks!
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