Expert: Scott A Wilson - 9/18/2010

Question
Sir, I have a few question regarding algebra imaginary
numbers.

1.)find the modules of:
                        (2+√-3)^2

2.)find the value of x and y:
                   (x+iy)^2=(5-12i)

3.)find the continued product of:
        (5-√-7)(5+√-7)(1-√-1)(1+√-1)

4.)for any complex number z,prove that:
     (1) |z|=|¯z|
      
     (2) |z|^2 = z ¯z

     (3) 1/2(z+¯z)) = Re(z)

     (4) 1/2i(z+¯z) = Im(z)

     (5) |z1 z2| = |z1|.|z2|

     (6) Re(z) <= |z|  
   p.s. I am not able to get the conjugate symbol over the
character z although it should be together.

Please answer my questions and provide explanation so that I
can understand the solution provided.

                                regards,
                                Prakash Kumar Singh


Answer
1.)find the modules of: (2+√-3)^2
The square is 2 + 2√-3 - 3 = -1 + 2√-3; I'm not sure what is meant in mathematics by modules.  From what I know, it is the division of a speech or story into sections.

2.)find the value of x and y: (x+iy)^2=(5-12i)
It is known that (x+iy)^2 = x²+2ixy-y².
Setting this to 5-21i means setting up the real to real and
imaginary to imaginary.  That is, x²-y²=5 and 2xy = -12.
The 2xy = -12 can be solved for x, and then that put into the 1st equation.  Once this has been done, use the quadratic equation to find x.  Once x is known, both equations should give the same value for y.

3.)find the continued product of: (5-√-7)(5+√-7)(1-√-1)(1+√-1)
That can be split into (5-√-7)(5+√-7) and (1-√-1)(1+√-1).
Remember that (a+b)(a-b) = a² - b²?  Well, that applies.
It is (25 - -7)(1 - -1) = 32*2.  You can finish that.

4.)for any complex number z,prove that:
(1) |z|=|¯z|  Let z = a + bi, so ¯z = a - bi;
|z| = √(a²+b²); |z¯| = √(a²+(-b)²) = √(a²+b²).

(2) |z|^2 = z ¯z; if z = a + bi, then ¯z = a - bi;
|z|² = a² + b²; z ¯z = (a+bi)(a-bi) = a² + b²;

(3) (z+¯z)/2 = Re(z); since ¯z has the negative of the imaginary part of z, the imaginary part cancels and the real part is both multiplied and divided by 2.

(4) i(z+¯z)/2 = Im(z); I'm not so sure about that
If z = a + bi, then ¯z = a - bi; this makes i(z+¯z)/2 = ia;
Perhaps it should be i(¯z-z)/2, and that would b

(5) |z1 z2| = |z1|.|z2|
When complex numbers are converted to polar notation, they give
(r,Θ) where r is the length and Θ is the angle.
If z1 = (r1,Θ1) and z2 = (r2,Θ2), then z1*z2 = (r1*r2, Θ1+Θ2).
The |z| = r, so the |r1*r2, Θ1+Θ2| = r1*r2, |z1| = r1, and |z2| = r2.
This says that r1*r2 = r1*r2.

(6) Re(z) <= |z| ; if z = a + bi, then Re(z) = a and |z| = √(a²+b²).
It is known that a = √a², so when b² is added to a², a greater result is gotten.