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secondary 2 | Maths
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complex numbers
Date Posted: 3 years ago
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Answer to you deleted previous problem:
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_12
https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_12
1.
If we let z = x + yi, then
a)
Reflection in the origin
= -x - yi
= -(x + yi)
= -z
It basically means the point is inverted, diagonally across the axes through the origin, such that both real and imaginary parts become their negatives. For example, 4 - 3i becomes -4 + 3i)
b)
Reflection in the x-axis means the real coordinate/part remains the same, but the imaginary part is now negative of the original.
So we would get x - yi.
This is known as the complex conjugate, z*
c)
Reflection in the y-axis means the imaginary coordinate/part remains the same, but the real part is now negative of the original.
So we would get -x + yi
= -(x - yi)
= -z*
As we can see, z and -z are diagonally opposite each other. Same for z* and -z*.
z and z* are directly above/below each other, same for -z and -z*
z and -z* are left/right of each other, same for -z and z*
If we let z = x + yi, then
a)
Reflection in the origin
= -x - yi
= -(x + yi)
= -z
It basically means the point is inverted, diagonally across the axes through the origin, such that both real and imaginary parts become their negatives. For example, 4 - 3i becomes -4 + 3i)
b)
Reflection in the x-axis means the real coordinate/part remains the same, but the imaginary part is now negative of the original.
So we would get x - yi.
This is known as the complex conjugate, z*
c)
Reflection in the y-axis means the imaginary coordinate/part remains the same, but the real part is now negative of the original.
So we would get -x + yi
= -(x - yi)
= -z*
As we can see, z and -z are diagonally opposite each other. Same for z* and -z*.
z and z* are directly above/below each other, same for -z and -z*
z and -z* are left/right of each other, same for -z and z*
If I have time, I will look at question 2.
See 1 Answer
f(z) = (a + bi)z , where a and b are positive numbers.
So the point is z and the image is (a + bi)z, another point.
①|a + bi| = 8 means that the modulus, a² + b² = 8² = 64
② b² = m/n
③ relatively prime is another term for coprime. So this means that the only number that divides both m and n exactly is 1.
Since the image is equidistant from z and the origin(0,0i), we just need to find the distance/length/modulus from origin to (a + bi)z, and from z to (a + bi)z, and equate them.
|(a+bi)z - 0 - 0i| = |(a + bi)z - z|
|(a+bi)z| = |z(a + bi - 1)|
|z||a+bi| = |z||a + bi - 1|
|a+bi| = |(a-1) + bi|
a² + b² = (a - 1)² + b²
a² + b² = a² - 2a + 1 + b²
2a = 1
a = ½
Now from ①, we know that a² + b² = 64
So b² = 64 - a²
b² = 64 - (½)²
m/n = 255/4 (from ② we know b² = m/n)
Since they are coprime, and we have gotten the simplest form for the fraction, then m = 255 and n = 4
So m + n = 255 + 4 = 259
So the point is z and the image is (a + bi)z, another point.
①|a + bi| = 8 means that the modulus, a² + b² = 8² = 64
② b² = m/n
③ relatively prime is another term for coprime. So this means that the only number that divides both m and n exactly is 1.
Since the image is equidistant from z and the origin(0,0i), we just need to find the distance/length/modulus from origin to (a + bi)z, and from z to (a + bi)z, and equate them.
|(a+bi)z - 0 - 0i| = |(a + bi)z - z|
|(a+bi)z| = |z(a + bi - 1)|
|z||a+bi| = |z||a + bi - 1|
|a+bi| = |(a-1) + bi|
a² + b² = (a - 1)² + b²
a² + b² = a² - 2a + 1 + b²
2a = 1
a = ½
Now from ①, we know that a² + b² = 64
So b² = 64 - a²
b² = 64 - (½)²
m/n = 255/4 (from ② we know b² = m/n)
Since they are coprime, and we have gotten the simplest form for the fraction, then m = 255 and n = 4
So m + n = 255 + 4 = 259
thank you!
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