Consider the complex number \(\omega = \frac{{z + {\text{i}}}}{{z + 2}}\), where \(z = x + {\text{i}}y\) and \({\text{i}} = \sqrt { - 1} \).

(a)     If \(\omega = {\text{i}}\), determine z in the form \(z = r\,{\text{cis}}\,\theta \).

(b)     Prove that \(\omega = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\).

(c)     Hence show that when \(\operatorname{Re} (\omega) = 1\) the points \((x,{\text{ }}y)\) lie on a straight line, \({l_1}\), and write down its gradient.

(d)     Given \(\arg (z) = \arg (\omega) = \frac{\pi }{4}\), find \(\left| z \right|\).

(a)     METHOD 1

\(\frac{{z + {\text{i}}}}{{z + 2}} = {\text{i}}\)

\(z + {\text{i}} = {\text{i}}z + 2{\text{i}}\)     M1

\((1 - {\text{i}})z = {\text{i}}\)     A1

\(z = \frac{{\text{i}}}{{1 - {\text{i}}}}\)     A1

EITHER

\(z = \frac{{{\text{cis}}\left( {\frac{\pi }{2}} \right)}}{{\sqrt 2 \,{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)}}\)     M1

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{4\pi }}{4}} \right)} \right)\)     A1A1

OR

\(z = \frac{{ - 1 + {\text{i}}}}{2}{\text{ }}\left( { = - \frac{1}{2} + \frac{1}{2}{\text{i}}} \right)\)     M1

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\)     A1A1

[6 marks]

METHOD 2

\({\text{i}} = \frac{{x + {\text{i}}(y + 1)}}{{x + 2 + {\text{i}}y}}\)     M1

\(x + {\text{i}}(y + 1) = - y + {\text{i}}(x + 2)\)     A1

\(x = - y;{\text{ }}x + 2 = y + 1\)     A1

solving, \(x = - \frac{1}{2};{\text{ }}y = \frac{1}{2}\)     A1

\(z = - \frac{1}{2} + \frac{1}{2}{\text{i}}\)

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\)     A1A1

Note: Award A1 fort the correct modulus and A1 for the correct argument, but the final answer must be in the form \({\text{r}}\,{\text{cis}}\,\theta \). Accept 135° for the argument.

 

[6 marks]

 

(b) substituting \(z = x + {\text{i}}y\) to obtain \(w = \frac{{x + (y + 1){\text{i}}}}{{(x + 2) + y{\text{i}}}}\)     (A1)

use of \((x + 2) - y{\text{i}}\) to rationalize the denominator     M1

\(\omega = \frac{{x(x + 2) + y(y + 1) + {\text{i}}\left( { - xy + (y + 1)(x + 2)} \right)}}{{{{(x + 2)}^2} + {y^2}}}\)     A1

\( = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\)     AG

[3 marks]

 

(c)     \(\operatorname{Re} \omega = \frac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}} = 1\)     M1

\( \Rightarrow {x^2} + 2x + {y^2} + y = {x^2} + 4x + 4 + {y^2}\)     A1

\( \Rightarrow y = 2x + 4\)     A1

which has gradient m = 2     A1

[4 marks]

 

(d)     EITHER

\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\)     (A1)

\(\omega = \frac{{2{x^2} + 3x}}{{{{(x + 2)}^2} + {x^2}}} + \frac{{{\text{i}}(3x + 2)}}{{{{(x + 2)}^2} + {x^2}}}\)

if \(\arg (\omega) = \theta \Rightarrow \tan \theta = \frac{{3x + 2}}{{2{x^2} + 3x}}\)     (M1)

\(\frac{{3x + 2}}{{2{x^2} + 3x}} = 1\)     M1A1

OR

\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\)     A1

\(\arg (w) = \frac{\pi }{4} \Rightarrow {x^2} + 2x + {y^2} + y = x + 2y + 2\)     M1

solve simultaneously     M1

\({x^2} + 2x + {x^2} + x = x + 2x + 2\) (or equivalent)     A1

THEN

\({x^2} = 1\)

\(x = 1{\text{ (as }}x > 0)\)     A1

Note: Award A0 for x = ±1.

 

\(\left| z \right| = \sqrt 2 \)     A1

Note: Allow FT from incorrect values of x.

 

[6 marks]

Total [19 marks]

Many candidates knew what had to be done in (a) but algebraic errors were fairly common. Parts (b) and (c) were well answered in general. Part (d), however, proved beyond many candidates who had no idea how to convert the given information into mathematical equations.