All categories

3 years ago

Find the sum of 9-4i and it’s complex conjugate

1 answer:

4 0

The complex conjugate of something is essentially just taking the opposite sign of the imaginary part. The following are complex conjugates:

a + bi

a - bi

In this case the complex conjugate of 9 - 4i is 9 + 4i

Now we must add them together like so:

(9 - 4i) + (9 + 4i)

9 - 4i + 9 + 4i

^^^Combine like terms

(9 + 9) + (-4i + 4i)

18 + 0

As you can see when you add complex conjugates your answer will be a real number

Hope this helped!

~Just a girl in love with Shawn Mendes

You might be interested in

Worth 15 POINTS!<br><br> Create a story for the equation <br> y= -2x + 14

Ierofanga [76]

-2x+14 is the same as 14 - 2x

Let's say you started off with $14. If you want to buy a soda worth $2, then you have 14-2 = 12 dollars left. If you buy two sodas, then you spend 2*2 = 4 dollars with 14-4 = 10 dollars left.

In general, buying x sodas will cost you 2*x dollars and you have 14 - 2x dollars left over. The x is simply a placeholder for a whole number. For example, if x = 3, then...

14 - 2x = 14 - 2*3 = 14 - 6 = 8

meaning buying 3 sodas cost you $6 and you have $8 left over.

If y is the amount left over, then we can say y = 14 - 2x which is equivalent to y = -2x+14

note: graphing this equation will go through the two points (0,14) and (7,0) as shown in the image below. A graph is handy to help see various points on the line. Each point represents the amount of sodas you can buy (x) and the amount left over in your pocket (y). Keep in mind that neither x nor y can be negative, so it only makes sense to restrict the graph.

3 0

3 years ago

$668 at 9.25% for 5 years

madreJ [45]

Answer:

$21.17

Step-by-step explanation:

Simple Interest=P(1+r)^t

Compounded Interest=P(e^rt)

SI=668*(1.0925)^5

SI=1039.64

CI=668(e^(0.0925*5))

CI=1060.81

1060.81-1039.64=21.17

So the difference is $21.17, thus doing compounded interest is better.

8 0

3 years ago

Please help meeee !!

katrin [286]

Answer:

The answer is B

3 0

3 years ago

5/5 + 2/5 show on a numbe<br> line

mixer [17]

I have a pic. I'll message you

4 0

3 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

3 years ago