How to create an inverse function in Lisp

[RusskiSoldat]

I'm using Dr. Scheme and need to translate the following function (given a + bi is an imaginary number) into Scheme:
a - bi / (a^2) + (b^2)

Here's what I already have:
(define complex (lambda (a b)
(cons a (cons b '()))))

(define getReal (lambda (n)
(car n)))

(define getImaginary (lambda (n)
(cadr n)))

(define x (complex 1 2))
(define y (complex 3 4))

(define add (lambda (x y)
(cons (+ (car x) (car y)) ( list (+ (cadr x) (cadr y))))))

(define addnice (lambda (x y)
(complex (+ (getReal x) (getReal y))
(+ (getImaginary x) (getImaginary y)))))


(define subtract (lambda (x y)
(cons (- (car x) (car y)) ( list (- (cadr x) (cadr y))))))

(define subtractnice (lambda (x y)
(complex (- (getReal x) (getReal y))
(- (getImaginary x) (getImaginary y)))))

(define multiply (lambda (x y)
(cons (- (* (car x) (car y)) (* (cadr x) (cadr y))) (list (+ (* (car x) (cadr y)) (* (car y) (cadr x)))))))

(define multiplynice (lambda (x y)
(complex (- (* (getReal x) (getReal y))
(getImaginary x) (getImaginary y))
(+ (* (getReal x) (getImaginary y))
(* (getImaginary x) (getReal y))))))

the "nice" versions of the functions are simply the easier way to do them.

[Erik]

You realize complex numbers are already implemented in Scheme?
You can simply write complex numbers like 2+5i or 6-8i and use them like any other number.
Or did you have to implement complex numbers for a school assignment?

I'm not sure what you're asking but a - bi / (a^2) + (b^2) translates simply into (/ (- a bi) (+ (* a a) (* b b)))