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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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