Let cis Θ = cos Θ + i sin Θ. Let z1 = 3 cis 25 and let z2 = 4 cis 115
Find z2^6
To find z2^6, we need to understand how to raise a complex number to a power.
First, let's express z2 in rectangular form. The rectangular form of a complex number is written as z = a + bi, where a and b are real numbers.
Given z2 = 4 cis 115, we can convert it to rectangular form using Euler's formula: cis(Θ) = cos(Θ) + i sin(Θ).
cos(115°) = -0.4226 (rounded to four decimal places)
sin(115°) = 0.9063 (rounded to four decimal places)
So z2 can be written as:
z2 = 4(cos(115°) + i sin(115°))
= 4(-0.4226 + 0.9063i)
Now, to raise z2 to the power of 6, we use De Moivre's theorem. It states that for any complex number z = r * cis Θ, z^n = r^n * cis (nΘ).
In our case, z2^6 = (4)^6 * cis(6 * 115°).
Calculating (4)^6:
(4)^6 = 4^6 = 4096
Calculating 6 * 115°:
6 * 115° = 690°
So z2^6 = 4096 * cis 690°.
Now, if we want to express z2^6 in rectangular form, we can convert it back using Euler's formula.
cos(690°) = cos(690° - 360°) = cos(330°) = -0.866 (rounded to three decimal places)
sin(690°) = sin(690° - 360°) = sin(330°) = -0.5
Therefore,
z2^6 = 4096 * (-0.866 - 0.5i)
Hence, z2^6 can be written as -4096.866 - 2048i.