Help meh
If z1=5+5i and Z2=8(cos(3pi/7)+ isin(3pi/7) then z1z2=

The "cis" is shorthand for "cos + i*sin"

Work Shown:

z1 = 5+5i is in the form a+bi with a = 5, b = 5

Use these values to find r and theta

r = sqrt(a^2+b^2)

r = sqrt(5^2+5^2)

r = sqrt(50)

r = sqrt(25*2)

r = sqrt(25)*sqrt(2)

r = 5sqrt(2)

theta = arctan(b/a)

theta = arctan(5/5)

theta = arctan(1)

theta = pi/4 radians

This tells us

z1 = 5+5i

turns into

z1 = 5sqrt(2)*(cos(pi/4)+i*sin(pi/4))

which is the polar form representation.

The notation "cos(x)+i*sin(x)" is cumbersome to write down. Common shorthand convention is to use "cis", the letters of which stand for the first letter of "cosine", then the "i", then the first letter of "sine" in that order.

In other words,

cos(x)+i*sin(x) = cis(x)

cos(pi/4)+i*sin(pi/4) = cis(pi/4)

and so on. It makes using the rule shown later much easier to work with.

What we really have are these two complex numbers

z1 = 5sqrt(2)*cis(pi/4)

z2 = 8*cis(3pi/7)

Now we'll use this rule

if z1 = r1*cis(theta1) and z2 = r2*cis(theta2)

then z1*z2 = r1*r2*cis(theta1+theta2)

we multiply the magnitudes r1 and r2; while adding the angles theta1 and theta2

So,

z1 = 5sqrt(2)*cis(pi/4)

z2 = 8*cis(3pi/7)

z1*z2 = 5sqrt(2)*8*cis(pi/4+3pi/7)

z1*z2 = 40sqrt(2)*cis(7pi/28+12pi/28)

z1*z2 = 40sqrt(2)*cis(19pi/28)

z1*z2 = 40sqrt(2)*( cos(19pi/28) + i*sin(19pi/28) )

The final r value is 40*sqrt(2)

The final theta value is 19pi/28 radians

To convert this into a+bi form, you would compute the sine and cosine values, distribute then simplify. I find it easier to keep it in polar form.

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.