Given z_1=2(cos pi/6+i sin pi/6) and z_2=3(cos pi/4 +i sin pi/4) find z_1z_2 where o<theta<2pi
AAAaannndd the bot gets it wrong yet again!
5pi/12 < pi, so it is in the desired range. So the answer is
6 cis(5pi/12)
I apologize for my mistake. You are correct that 5pi/12 is within the desired range of 0 < theta < 2pi. The correct answer is indeed 6cis(5pi/12). Thank you for bringing this to my attention.
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To find the product of two complex numbers, z_1 and z_2, we can multiply their magnitudes and add their arguments.
In this case, we have:
z_1 = 2(cos(pi/6) + i sin(pi/6))
z_2 = 3(cos(pi/4) + i sin(pi/4))
First, let's find the magnitude of z_1 and z_2:
|z_1| = √(Re[z_1]^2 + Im[z_1]^2)
= √(2^2 + 0^2) = 2
|z_2| = √(Re[z_2]^2 + Im[z_2]^2)
= √(3^2 + 0^2) = 3
Next, let's find the argument of z_1 and z_2:
arg(z_1) = atan2(Im[z_1], Re[z_1])
= atan2(sin(pi/6), cos(pi/6))
= pi/6
arg(z_2) = atan2(Im[z_2], Re[z_2])
= atan2(sin(pi/4), cos(pi/4))
= pi/4
The product of z_1 and z_2 can be found by multiplying their magnitudes and adding their arguments:
z_1z_2 = |z_1||z_2| * exp(i(arg(z_1) + arg(z_2)))
= 2 * 3 * exp(i(pi/6 + pi/4))
Now, we need to add the arguments inside the exponential function:
arg(z_1z_2) = pi/6 + pi/4
To keep the angle between 0 and 2pi, we can convert the sum to a common denominator:
arg(z_1z_2) = (2pi/12 + 3pi/12) = 5pi/12
Finally, we can rewrite z_1z_2 in trigonometric form:
z_1z_2 = 6 * (cos(5pi/12) + i sin(5pi/12))
Therefore, z_1z_2 = 6(cos(5pi/12) + i sin(5pi/12)).
Using the polar form of complex numbers, we can write:
z_1 = 2 cis(pi/6)
z_2 = 3 cis(pi/4)
To find z_1z_2, we multiply the magnitudes and add the angles:
z_1z_2 = (2*3) cis(pi/6 + pi/4)
= 6 cis(5pi/12)
However, we need to express this answer in the given range of angles, which is 0 < theta < 2pi. We notice that 5pi/12 is greater than pi and less than 3pi/2, so we can add 2pi to bring it back into the desired range:
z_1z_2 = 6 cis(5pi/12 + 2pi)
= 6 cis(29pi/12)
= 6 cis(pi/12)
Therefore, z_1z_2 = 6(cos pi/12 + i sin pi/12).