Answer:
C. 56 is the correct answer, my bad.
Step-by-step explanation:
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.
![(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4](https://tex.z-dn.net/?f=%28z_1%29%5E1%5E2%3D%5B4e%5E%7B-i%5Cfrac%20%5Cpi6%7D%5D%5E1%5E2%3D4%5E1%5E2%5Ccdot%20%28e%5E%7B-i%5Cfrac%20%5Cpi6%7D%29%5E1%5E2%3D2%5E2%5E4%5Ccdot%20e%5E%7B-i%2812%29%5Cfrac%5Cpi6%7D%5C%5C%3D2%5E2%5E4%20e%5E%7B-i%5Ccdot2%5Cpi%7D%3D2%5E2%5E4)
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.
Answer: 13
Step-by-step explanation:
I used Pythagorean theorem to solve this problem. As DC and AD is 12 and 5, you would do 12²+5²= c². 12²= 144 and 5²= 25. 144+25= 169. It doesn't end there. Do the square root of 169 ---> √169=13.
To find the solution, you plot 5 as the y intercept. Now x has a slope of -1x/3 so you move down 1 and over 3. If it is the greater than sign has a line underneath it, you use a solid line.If there is no line, use a dotted line Since the arrow is pointing towards x, you will shape upwards because it is either greater or equal.
