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Roots of a Complex Equation


Date: 04/28/97 at 01:44:57
From: raylim
Subject: Complex number

The following question is from the British A Level exam. Can you solve 
it for me? I couldn't get the solution because I had no idea how to 
begin.

You are given that cos t + isin t is the root of z^3 + az + b = 0 in 
which sin t is not equal to 0. If a and b are real and not equal to 
zero, show that a + b^2 = 1.

Thanks,
Raylim


Date: 04/28/97 at 16:25:59
From: Doctor Anthony
Subject: Re: Complex number

Raylim,

If z = cos(t) + i.sin(t) is a root, then since the coefficients of the 
equation are real, it is also true that z = cos(t) - i.sin(t) is a 
root.

It follows that [z-cos(t)-i.sin(t)] and [z-cos(t)+i.sin(t)] are 
factors:
              = (z-cos(t))^2 + sin^2(t)  is a factor

              =  z^2 - 2z.cos(t) + cos^2(t) + sin^2(t)  is a factor

              =  z^2 - 2z.cos(t) + 1    is a factor

If we do long division of this expression into z^3 + az + b, it must 
go in exactly with a remainder of 0.

In fact, the quotient is z + 2.cos(t) and the remainder is:

      Remainder = z[a-1+4cos^2(t)] + [b - 2.cos(t)] = 0

We can put each part of this expression separately equal to 0 since 
one part is multiplied by z (meaning that is also multiplied by i and 
so is imaginary) and the other part, b-2.cos(t), is real:

 b-2.cos(t) = 0  
          b = 2.cos(t)    
        b^2 = 4.cos^2(t)

Also:   a-1 + 4.cos^2(t) = 0

        a-1 + b^2 = 0

          a + b^2 = 1  

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 04/28/97 at 16:45:55
From: Doctor Wilkinson
Subject: Re: Complex number

Raylim,

You are given that one of the roots is of the form cos t + i sin t, 
where sin t is different from 0.  In particular, this is a complex 
root, and you are probably expected to know that the complex roots of 
an equation with real coefficients occur in conjugate pairs, so that a 
second root is cos t - i sin t.  The product of these two roots is 
cos^ t + sin^ t, which is 1.

Now we can forget about the trigonometry and call the complex roots u 
and v, where uv = 1.  The remaining root must be real, and we'll call 
it w.

Now we can write:

 z^3 + az + b = (z - u) (z - v)(z - w)

Now multiply out the right side.  This gives:

 a^3 - (u + v + w) z^2 + (uv + uw +vw) z - uvw

This must be identically equal to the left side, so we must have:

 u + v + w = 0

 uv + uw +vw = a

 uvw = -b

Now we already now that uv = 1, so we can simplify these to:

 (u + v) + w = 0
 
 1 + (u + v) w = a

 w = -b

The first equation gives us u + v = -w.  Substitute this into the 
second equation to get:

 1 - w^2 = a

But the third equation says w = -b, so:

 1 - b^2 = a or a + b^2 = 1
 
-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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