I have a problem about complex number.
Let z=7e^(7*pi*i/4) .
Write z, z^4, z^7, z^(-2) in the form a + b i
I have no idea how to do it. Any hints will be appreciated.
z=7e^(7*pi*i/4)
z^4 = 7^4 e^(7*pi*i)
And then you use that
e^(i theta) = cos(theta) + i sin(theta)
To write complex numbers in the form a + bi, we need to convert them from exponential form to rectangular form. In the given question, we have z = 7e^(7πi/4).
To convert this to rectangular form, we can use Euler's formula: e^(ix) = cos(x) + i*sin(x). In this case, we have z = 7e^(7πi/4) = 7(cos(7π/4) + i*sin(7π/4)).
Now let's solve it step by step:
1. Writing z in a + bi form:
Since cos(7π/4) = -sqrt(2)/2 and sin(7π/4) = sqrt(2)/2, we can substitute these values in the equation:
z = 7(-sqrt(2)/2 + i * sqrt(2)/2)
Expanding, we get:
z = -7sqrt(2)/2 + 7i sqrt(2)/2
So, z = -7sqrt(2)/2 + 7i sqrt(2)/2 in the form a + bi.
2. Writing z^4 in a + bi form:
To find z^4, we raise z to the power of 4:
z^4 = (7(-sqrt(2)/2 + i * sqrt(2)/2))^4
Expanding and simplifying, we get:
z^4 = 7^4 * (sqrt(2)/2 - i * sqrt(2)/2)^4
Note that (sqrt(2)/2 - i * sqrt(2)/2)^4 can be simplified as follows:
(sqrt(2)/2 - i * sqrt(2)/2)^4 = (1/2^(1/2) * (1 - i))^4
Applying the binomial theorem, we can expand this expression:
(1/2^(1/2) * (1 - i))^4 = (1/2 * (1 - i))^4
Expanding and simplifying, we get:
(1/2 * (1 - i))^4 = 1/16 * (1 - 4i + 6 - 4i + 1)
= 1/16 * (8 - 8i)
= 1/2 - 1/2i
So, z^4 = 1/2 - 1/2i in the form a + bi.
3. Writing z^7 in a + bi form:
To find z^7, we raise z to the power of 7:
z^7 = (7(-sqrt(2)/2 + i * sqrt(2)/2))^7
Expanding and simplifying, we get:
z^7 = 7^7 * (sqrt(2)/2 - i * sqrt(2)/2)^7
Using the same process as before, we can simplify (sqrt(2)/2 - i * sqrt(2)/2)^7:
(sqrt(2)/2 - i * sqrt(2)/2)^7 = (1/2 * (1 - i))^7
Expanding and simplifying, we get:
(1/2 * (1 - i))^7 = 1/128 * (1 - 7i)
So, z^7 = 1/128 - 7/128i in the form a + bi.
4. Writing z^(-2) in a + bi form:
To find z^(-2), we take the reciprocal of z^2:
z^(-2) = 1/z^2
To simplify this, we need to find the reciprocal of z:
1/z = 1/(7(-sqrt(2)/2 + i * sqrt(2)/2))
Multiplying the numerator and denominator by the conjugate of the denominator, we get:
1/z = (1/(7(-sqrt(2)/2 + i * sqrt(2)/2)))) * (-sqrt(2)/2 - i * sqrt(2)/2)) / (-sqrt(2)/2 - i * sqrt(2)/2))
Simplifying, we get:
1/z = -sqrt(2)/2 - i * sqrt(2)/2
Now, to find z^(-2), we square this reciprocal:
z^(-2) = (-sqrt(2)/2 - i * sqrt(2)/2)^2
Expanding and simplifying, we get:
z^(-2) = 1/2 + 1/2i
So, z^(-2) = 1/2 + 1/2i in the form a + bi.
To summarize:
z = -7sqrt(2)/2 + 7i sqrt(2)/2
z^4 = 1/2 - 1/2i
z^7 = 1/128 - 7/128i
z^(-2) = 1/2 + 1/2i