Determine the imaginary part of complex number z if z^2=3+4i .

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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