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(ii) The domain and range of f are both R.

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(iii) The domain is[5,) and the range is [0,).

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(iv) Since we must have 1 x2 > 0, the domain of f(x) is the interval (1, 1). The rangeis(, 0].

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5. (i) The natural domain of f is C. Every non-zero complex number has five 5th roots andthe number 0 has one 5th root, namely 0; hence every complex number is the result of

raising at least one complex number to the power 5. Therefore the range is C.

(ii) Consider z C with z= 0. Then z = rei for some real r and with r> 0, andf(z) = 1

z = 1

rei. However, 1

z is not defined when z= 0; hence the natural domain of

f is C

\ {0

}. Since w= 1

zif and only ifz= 1

w, it follows that the range is also C

\ {0

}.

(iii) The complex exponential function has natural domain C; hence f has natural domainC. Ifz= x +iy with x, yR then ez =exeiy, and|ez|=|ex| |eiy|= ex, since ex is realand positive and|eiy|= (cosy)2 + (siny)2 = 1. Since ex can take any positive value,f has range(0,).

(iv) The natural domain is C. Since|z| is real and nonnegative, and since every nonnegativereal number arises as|z| for some z C, the range of f is the same as the range of thereal exponential function on the interval [0,), namely [1,). (Recall that e0 =1.)

6. By de Moivre,(cos+i sin )5 =cos 5+i sin5, and by the binomial theorem

(cos+i sin )5 =cos5 +5 cos4 (i sin) +10 cos3 (i2 sin2 )+10 cos2 (i3 sin3 ) +5 cos(i4 sin4 ) + (i5 sin5 )

=cos5 10 cos3 sin2 +5 cos sin4 +i(5cos4 sin 10 cos2 sin3 +sin5 ).

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Equating real and imaginary parts of the two expressions for (cos+i sin)5 gives

cos5=cos5 10 cos3 sin2 +5 cossin4 and

sin5=5 cos4 sin 10 cos2 sin3 +sin5 .

7. (i) Write z= x +iy where x, y are real. Then e

z

=ex+iy

=ex

eiy

, and since y is real and ex

is real and positive it follows that ex is the modulus of ez and y is (one of the valuesof) the argument of ez. In Question 1 we found that

3 i in polar exponential form

is 2ei/6. So the equation becomes

exeiy =2ei/6 for some x, y R,

and, equating the moduli and arguments of the two sides, this is equivalent to ex = 2and y =

6 +2k for some k Z. So z is a solution of ez = 3 i if and only if

z= ln 2+i(6

+2k) for some k Z. (So there are infinitely many solutions.)(ii) Write z = x+ iy where x, y

R. Since

1 in polar exponential form is 1ei, the

equation becomes

exeiy =1ei for some x, yRwhich is equivalent to ex =1 and y= + 2kfor some k Z. But ex =1 if and onlyifx= 0; so z is a solution ofez =1 if and only ifz= i(2k+1) for some k Z.

(iii) The question asks us to find the two square roots of ei/4. Writing z = rei (withr, R and r> 0), the equation becomes r2e2i = ei/4, which is is satisfied if andonly if r2 = 1 and 2 =

4 +2k for some k Z. So r = 1 and =

8 +k for

some k Z. Since values of that differ by a multiple of 2 give the same value ofei, there are exactly two distinct solutions for z, corresponding to even values ofkand

odd values ofk. The values ofk that give the principal arguments of the two solutionsare in fact k= 0 and k=1. The solutions are z = ei(/8) (corresponding to k= 0)z= ei(7/8) (corresponding to k=1).

8. The binomial theorem withn= 5 gives

sin5 =

12i

(ei ei)5= 1

32i(e5i 5e4iei +10e3ie2i 10e2ie3i +5eie4i e5i)

= 132i

((e5i e5i) 5(e3i e3i) +10(ei ei))= 1

32(2sin5

10sin3+20 sin ).

Therefore

0sin5 d=

1

32

0(2sin5 10sin3+20 sin ) d

= 132

( 25

[cos5]0 + 10

3[cos3]0 20[cos]0 )

= 132

( 45 20

3 +40)

= 1615

.

9. (i) Since z= 0+2i we have ez =e0e2i =1(cos2+i sin2) 0.41+0.91i.

(ii) We have ez

=e4

(cos+ i sin) =e4

.(iii) This time ez =ecos ei sin =ecos (cos(sin) +i sin(sin )).

10. (i) f(0) =2 02 +3 0 4=4.(ii) f(2) =2 22 +3 2 4= 8+6 4= 10.

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(iii) f(

2) =2(

2)2 +3

2 4= 32.(iv) f(1+

2) =2(1+

2)2 +3(1+

2) 4= 2(3+22) +3+32 4= 5+72.

(v) f(x) =2(x)2 +3(x) 4= 2x2 3x 4.(vi) f(x+1) =2(x+1)2 +3(x+1) 4= (2x2 +4x+2) + (3x+3) 4= 2x2 +7x+1.(vii) 2f(x) =2(2x2 +3x 4) =4x2 +6x 8.(viii) f(2x) =2(2x)2 +3(2x)

4= 8x2 +6x

4.

11. We find that (fg)(x) = f(g(x)) = sin(1 x), and (given that x is real) this is defined ifand only ifx0. So the largest possible domain for fg is [0,).Similarly, (gf)(x) = g(f(x)) = 1 sinx, which is defined if and only if sinx 0. Sincesin is positive in the first and second quadrants, this condition holds if and only ifx has the

form x0+ 2k for some integer k and some x0 in the interval [0, ]. So the largest possibledomain forgf is{x R |2k x(2k+ 1) for some k Z }, the union of all the closedintervals[2k, (2k+1)] for k Z.

12. (i) We may calculate sin |x| for any x R; so the natural domain of f is R. Since1 sin 1 for all R, it is certainly true that f(x) = sin |x| [1, 1] forall x. So the range of f is a subset of [1, 1]. Furthermore, from the graph of sinxit can be seen that sinx takes on all values between sin

2 = 1 and sin 3

2 =1 as x

increases from 2

to 32

. In particular, for each y[1, 1] there is an x[ 2, 3

2 ] such

that f(x) = sin |x| = sinx = y. So each element of [1, 1] is in the range of f, whichtherefore is the whole interval [1, 1].

(ii) The natural domain ofx+1 is [1,), and since the real exponential function can

take any input, this is also the domain ofg. As x moves from1 towards , the valuesof

x+1 increase continuously, starting at 0 and tending to , and the values ofe

x+1

increase continuously, starting at e0 =1 and tending to . So the range ofg is [1,).

We can expand this argument, and make it more explicit, as follows. If x [1,)then

1+x [0,), and e

1+x [1,) since e0 is 1 and et increases as t increases.

So the range ofg is contained in [1,). And if y is any number in the interval [1,)then lny is a positive number such that y= elny, and if we put x= (lny)2 1 then wefind that

x+1 = lny and g(x) = e

x+1 = elny = y. So all numbers y[1,) are in

the range ofg.

13. (i) The domain of f is R = (,) and its range is [0,).(ii) The domain of f is R = (,) and its range is (1,).(iii) We know from lectures that the range ofez is C

\ {0

}. Hence every z

C can be used

as an input for f (since the denominator will never be zero). Thus the natural domain off is C. (Note that we could have written f(z) = ez as the definition of the function.)The range of fis the same as the range ofez, namely C \ {0}.

14. (i) The graphs of both f and g are hyperbolas, although on one branch of each hyperbolaa point is missing from the graph (because we have chosen not to include 0 in thedomain of fand not to include 1 in the domain ofg). The graphs of these hyperbolasshow that the ranges of both f and g are R \ {0, 1}: the hyperbola corresponding to fhas a horizontal asymptote at y= 0 and is missing the one point at which the functionvalue would have been 1 (since we have excluded x = 0) and similarly the hyperbolacorresponding to g has a horizontal asymptote at y = 1 and the one point that wouldhave given a function value of 0 has been excluded.

(ii) The formulas for the composite functions fg and gf are

(fg)(x) = 1

1 g(x) = 1

1 (x 1)/x = x

x (x 1) =x,

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and

(gf)(x) = f(x) 1

f(x) =

1/(1 x) 11/(1 x) =

1 (1 x)1

=x.

Hence, f and g undo each other; that is, they are mutually inverse functions. Inverse

functions will be studied in more detail later in the course.