What’s equivalent to z+(z+6)
2 answers:
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Answer and step-by-step explanation:
The polar form of a complex number is the number
where
is called the modulus and
is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.
(a) let's first calculate moduli and arguments
now we can write the two numbers as
(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:
(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.
Now, in the last step I've used the fact that , or in other words, the complex exponential is periodic with
as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.
Notation
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
. For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets. ![[( x^{3}+5x)]-[3 x^{2} -15]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D)
Let's next factor what is in each bracket:
![[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D%20%3D%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D)
Notice that both brackets have the same expression in them so now we factor that out:![[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)](https://tex.z-dn.net/?f=%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D%20%3D%20%28x-3%29%28%20x%5E%7B2%7D%20%2B5%29)
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get:
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
Let's next factor what is in each bracket:
Notice that both brackets have the same expression in them so now we factor that out:
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: