(please do it fast i need help its due is in 10 min) If A(2,2),B(8,6),P(4,2),Q(10,6) are any four points then prove that AB = PQ in length
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Answer:
Complex multiplication is a more difficult operation to understand from either an algebraic or a geometric point of view. Let’s do it algebraically first, and let’s take specific complex numbers to multiply, say 3 + 2i and 1 + 4i. Each has two terms, so when we multiply them, we’ll get four terms:
(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i2.
Now the 12i + 2i simplifies to 14i, of course. What about the 8i2? Remember we introduced me as an abbreviation for √–1, the square root of –1. In other words, i is something whose square is –1. Thus, 8i2 equals –8. Therefore, the product (3 + 2i)(1 + 4i) equals –5 + 14i.
If you generalize this example, you’ll get the general rule for multiplication
(x+yi)(u+vi) = (xu=yv)+(xv+yu)i
Remember that (xu – yv), the real part of the product, is the product of the real parts minus the product of the imaginary parts, but (xv + yu), the imaginary part of the product, is the sum of the two products of one real part and the other imaginary part.
Let’s look at some special cases of multiplication.
Multiplying a complex number by a real number
In the above formula for multiplication, if v is zero, then you get a formula for multiplying a complex number x + yi and a real number u together:
(x + yi) u = xu + yu i.
In other words, you just multiply both parts of the complex number by the real number. For example, 2 times 3 + i is just 6 + 2i. Geometrically, when you double a complex number, just double the distance from the origin, 0. Similarly, when you multiply a complex number z by 1/2, the result will be half way between 0 and z. You can think of multiplication by 2 as a transformation which stretches the complex plane C by a factor of 2 away from 0; and multiplication by 1/2 as a transformation which squeezes C toward 0.
Multiplication and absolute value.
Even though we’ve only done one case for multiplication, it’s enough to suggest that the absolute value of zw (i.e., distance from 0 to zw) might be the absolute value of z times the absolute value of w. It was when w was the real number u just above. In fact, this is true in general:
|zw| = |z| |w|
The verification of this identity is an exercise in algebra. In order to prove it, we’ll prove it’s true for the squares so we don’t have to deal with square roots. We’ll show |zw|2 = |z|2|w|2. Let z be x + yi, and let w be u + vi. Then, according to the formula for multiplication, zw equals (xu – yv) + (xv + yu)i. Recall from the section on absolute values that
|z|2 = x2 + y2
Similarly, we have
|w|2 = u2 + v2
and, since zw = (xu – yv) + (xv + yu)i,
|wz|2 = (xu – yv)2 + (xv + yu)2
So, in order to show |zw|2 = |z|2|w|2, all you have to do is show that
(xu – yv)2 + (xv + yu)2 = (x2 + y2) (u2 + v2)
and that’s a straightforward exercize in algebra.