20 POINTS Using mathematically precise language, explain in detail how you would multiply the complex number
z1=r1 (cos theta1+isin theta 1)
with the complex number
z2=r2 (cos theta 2+isin theta 2)

r-radius/radii
m-modulus/moduli
theta-argument represent angle theta

PRODUCT OF 2 COMPLEX NUMBERS: z1z2 = r1r2[cos (θ1+θ2) + isin (θ1+θ2)]

proof: r1(cos θ1 + isin θ1) × r2(cos θ2 + isin θ2)

= r1r2(cos θ1 + isin θ1) (cos θ2 + isin θ2)

= r1r2(cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 + i2

sin θ1 sin θ2)

= r1r2(cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 + (-1)sin θ1 sin θ2)

= r1r2(cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 - sin θ1 sin θ2)

Step-by-step explanation:

when multipying these variables, you need to take into accound that the number has a value greater that its reciprical. useing an algebratic calculator can help. my tutor explained that to me. hope it helps :D

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

a and c

step-by-step explanation: