If z is a non-zero complex number, we define \(L(z)\) by the equation

\[L(z) = \ln \left| z \right| + {\text{i}}\arg (z),{\text{ }}0 \leqslant \arg (z) < 2\pi .\]

(a)     Show that when z is a positive real number, \(L(z) = \ln z\) .

(b)     Use the equation to calculate

(i)     \(L( - 1)\) ;

(ii)     \(L(1 - {\text{i}})\) ;

(iii)     \(L( - 1 + {\text{i}})\) .

(c)     Hence show that the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\) does not hold for all values of \({z_1}\) and \({z_2}\) .

Let f be a function with domain \(\mathbb{R}\) that satisfies the conditions,

\(f(x + y) = f(x)f(y)\) , for all x and y and \(f(0) \ne 0\) .

(a)     Show that \(f(0) = 1\).

(b)     Prove that \(f(x) \ne 0\) , for all \(x \in \mathbb{R}\) .

(c)     Assuming that \(f'(x)\) exists for all \(x \in \mathbb{R}\) , use the definition of derivative to show that \(f(x)\) satisfies the differential equation \(f'(x) = k{\text{ }}f(x)\) , where \(k = f'(0)\) .

(d)     Solve the differential equation to find an expression for \(f(x)\) .

(a)     \(\left| z \right| = z\) , \(\arg (z) = 0\)     A1A1

so \(L(z) = \ln z\)     AG     N0

[2 marks]

 

(b)     (i)     \(L( - 1) = \ln 1 + {\text{i}}\pi = {\text{i}}\pi \)     A1A1     N2

(ii)     \(L(1 - {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{7\pi }}{4}\)     A1A1     N2

(iii)     \(L( - 1 + {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{3\pi }}{4}\)     A1     N1

[5 marks]

 

(c)     for comparing the product of two of the above results with the third     M1

for stating the result \( - 1 + {\text{i}} = - 1 \times (1 - {\text{i}})\) and \(L( - 1 + {\text{i}}) \ne L( - 1) + L(1 - {\text{i}})\)     R1

hence, the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\)

does not hold for all values of \({z_1}\) and \({z_2}\)     AG     N0

[2 marks]

Total [9 marks]

(a)     from \(f(x + y) = f(x)f(y)\)

for x = y = 0     M1

we have \(f(0 + 0) = f(0)f(0) \Leftrightarrow f(0) = {\left( {f(0)} \right)^2}\)     A1

as \(f(0) \ne 0\), this implies that \(f(0) = 1\)     R1AG     N0

[3 marks]

 

(b)     METHOD 1

from \(f(x + y) = f(x)f(y)\)

for y = –x , we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\)     M1A1

as \(f(0) \ne 0\) this implies that \(f(x) \ne 0\)     R1AG     N0

METHOD 2

suppose that, for a value of x, \(f(x) = 0\)     M1

from \(f(x + y) = f(x)f(y)\)

for \(y = - x\), we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\)     A1

substituting \(f(x)\) by 0 gives \(f(0) = 0\) which contradicts part (a)     R1

therefore \(f(x) \ne 0\) for all x.     AG     N0

[3 marks]

 

(c)     by the definition of derivative

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\)     (M1)

\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x)f(h) - f(x)f(0)}}{h}} \right)\)     A1(A1)

\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(h) - f(0)}}{h}} \right)f(x)\)     A1

\( = f'(0)f(x)\,\,\,\,\,\left( { = k{\text{ }}f(x)} \right)\)     AG     N0

[4 marks]

 

(d)     \(\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \int {k{\text{d}}x \Rightarrow \ln f(x) = kx + C} } \)     M1A1

\(\ln f(0) = C \Rightarrow C = 0\)     A1

\(f(x) = {{\text{e}}^{kx}}\)     A1     N1

Note: Award M1A0A0A0 if no arbitrary constant C .

 

[4 marks]

Total [14 marks]

Part A was answered well by a fair amount of candidates, with some making mistakes in calculating the arguments of complex numbers, as well as careless mistakes in finding the products of complex numbers.

Part B proved demanding for most candidates, particularly parts (c) and (d). A surprising number of candidates did not seem to know what was meant by the ‘definition of derivative’ in part (c) as they attempted to use quotient rule rather than first principles.