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### Complex Equations

```
Date: 6/14/96 at 10:44:57
From: Kees van Schaik
Subject: COMPLEX-question

Dear Doctor,

I have a little problem:

Let z be an element of the complex numbers. Then solve:

(1) cos z = 2
(2) z^2 - 2z = i
```

```
Date: 6/14/96 at 12:0:1
From: Doctor Anthony
Subject: Re: COMPLEX-question

(1) cos(z) = 2.  We have cos(z) = (e^(iz)+e^(-iz))/2 = 2.

So e^(iz)+e^(-iz) = 4.

Multiply through by e^(iz) and we get

e^(2iz) + 1 = 4e^(iz).

So e^(2iz) - 4e^(iz) + 1 = 0.  This is a quadratic in e^(iz)

e^(iz) = {4 +or- sqrt(16-4)}/2  = {4 +or- 2sqrt(3)}/2

= 2 +or- sqrt(3)

iz = ln(2 +or- sqrt(3))

z = -i*ln(2 +or- sqrt(3)) = i*ln{1/(2 +or- sqrt(3))}

= i*ln(2-sqrt(3))  or i*ln(2+sqrt(3))

(2) Solve  z^2-2z = i

We could complete the square getting z^2 - 2z + 1 = 1 + i.
Then (z - 1)^2 = 1 + i  and taking square roots

z-1 = +or- sqrt(1+i).

The square root of 1+i is more easily done if we note that

1 + i = sqrt(2)*e^(i*pi/4)

and +or- sqrt(1+i) = +or- 2^(1/4)*e^(i*pi/8).

And so finally z = 1 +or- 2^(1/4)*e^(i*pi/8)

z = 1 +or- 2^(1/4){cos(pi/8)+i*sin(pi/8)}.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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