Determine the imaginary part of complex number z if z^2=3+4i .
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We have to determine the imaginary part of the complex number z for z^2 = 3 + 4i
Let z = x + yi
z^2 = x^2 + y^2*i^2 + 2xyi = 3+ 4i
=> x^2 - y^2 + 2xyi = 3 + 4i
equate the real and imaginary coefficients
we get x^2 - y^2 = 3 and 2xy = 4 or xy = 2 or x = 2/y
substitute x = 2/y in x^2 - y^2 = 3
=> 4/y^2 - y^2 = 3
=> 4 - y^4 = 3y^2
=> y^4 + 3y^2 - 4 = 0
=> y^4 + 4y^2 - y^2 - 4= 0
=> y^2( y^2 + 4) -1 ( y^2 + 4) = 0
=> (y^2 - 1)(y^2 + 4) = 0
=> y^2 = 1 and y^2 = -4
but y is a real coefficient, so we ignore y^2 = -4
y^2 = 1
=> y = 1
Therefore the imaginary part of the complex number z is 1.
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First, we'll have to determine the complex number.
We'll write the rectangular form of a complex number:
z = a + bi
We'll raise to square both sides:
z^2 = (a+bi)^2
z^2 = a^2 + 2abi + b^2*i^2, but i^2 =-1
z^2 = a^2 + 2abi - b^2
From enunciation, we know that:
z^2 = 3+4i
Comparing, we'll get:
a^2 + 2abi - b^2 = 3 + 4i
a^2 - b^2 = 3 (1)
2ab = 4
ab = 2
b = 2/a (2)
We'll substitute (2) in (1):
a^2 - 4/a^2 = 3
We'll multiply by a^2 all over:
a^4 - 3a^2 - 4 = 0
We'll substitute a^2 = t
t^2 - 3t - 4 = 0
We'll apply quadratic formula:
t1 = [3 + sqrt(9 + 16)]/2
t1 = (3+5)/2
t1 = 4
a^2 = 4
a1 = 2 and a2 = -2
b1 = 2/a1 = 2/2 = 1
b2 = -1
t2 = (3-5)/2
t2 = -1
a^2 = -1
a3 = i and a4 = -i
b3 = 1/i = i/i^2 = -i
b4 = i
The complex number are:
z1 = 2-i and z2=-2+i
z3= i - i^2
z3 = 1 + i
z4 = -i + i^2
z4 = -1 - i
The imaginary parts of z are: Im(z) = {-1 ; 1}.
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