Express each of the complex number given in the Exercises 1 to 10 in the form \( a + ib \).
Q1
\( (5i) \left( -\frac{3}{5}i \right) \)
Answer:
\( (5i) \left( -\frac{3}{5}i \right) = -3i^2 \)
Since \( i^2 = -1 \), we get:
\( -3(-1) = 3 \)
Standard form: \( 3 + 0i \)
Q2
\( i^9 + i^{19} \)
Answer:
We know \( i^4 = 1 \).
\( i^9 = (i^4)^2 \cdot i = 1 \cdot i = i \)
\( i^{19} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i \)
\( i^9 + i^{19} = i - i = 0 \)
Standard form: \( 0 + 0i \)
Q3
\( i^{-39} \)
Answer:
\( i^{-39} = \frac{1}{i^{39}} = \frac{1}{(i^4)^9 \cdot i^3} = \frac{1}{1 \cdot (-i)} = \frac{1}{-i} \)
Multiply numerator and denominator by \( i \):
\( \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i \)
Standard form: \( 0 + i \)
Q4
\( 3(7 + i7) + i(7 + i7) \)
Answer:
\( = 21 + 21i + 7i + 7i^2 \)
\( = 21 + 28i + 7(-1) \)
\( = 21 - 7 + 28i \)
Standard form: \( 14 + 28i \)
Q5
\( (1 - i) - (-1 + i6) \)
Answer:
\( = 1 - i + 1 - 6i \)
\( = (1 + 1) + (-i - 6i) \)
Standard form: \( 2 - 7i \)
Q6
\( \left( \frac{1}{5} + i\frac{2}{5} \right) - \left( 4 + i\frac{5}{2} \right) \)
Answer:
Group real and imaginary parts:
\( = \left( \frac{1}{5} - 4 \right) + i \left( \frac{2}{5} - \frac{5}{2} \right) \)
\( = \left( \frac{1-20}{5} \right) + i \left( \frac{4-25}{10} \right) \)
Standard form: \( -\frac{19}{5} - i\frac{21}{10} \)
Q7
\( \left[ \left( \frac{1}{3} + i\frac{7}{3} \right) + \left( 4 + i\frac{1}{3} \right) \right] - \left( -\frac{4}{3} + i \right) \)
Answer:
First bracket:
\( \left( \frac{1}{3} + 4 \right) + i\left( \frac{7}{3} + \frac{1}{3} \right) = \frac{13}{3} + i\frac{8}{3} \)
Now subtract the last term:
\( \left( \frac{13}{3} + i\frac{8}{3} \right) - \left( -\frac{4}{3} + i \right) \)
\( = \left( \frac{13}{3} + \frac{4}{3} \right) + i\left( \frac{8}{3} - 1 \right) \)
\( = \frac{17}{3} + i\frac{5}{3} \)
Standard form: \( \frac{17}{3} + i\frac{5}{3} \)
Q8
\( (1 - i)^4 \)
Answer:
We can write \( (1 - i)^4 = [(1 - i)^2]^2 \).
\( (1 - i)^2 = 1^2 - 2i + i^2 = 1 - 2i - 1 = -2i \)
Now square the result:
\( (-2i)^2 = 4i^2 = 4(-1) = -4 \)
Standard form: \( -4 + 0i \)
Q9
\( \left( \frac{1}{3} + 3i \right)^3 \)
Answer:
Using \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \):
\( = (\frac{1}{3})^3 + 3(\frac{1}{3})^2(3i) + 3(\frac{1}{3})(3i)^2 + (3i)^3 \)
\( = \frac{1}{27} + 3(\frac{1}{9})(3i) + (9i^2) + 27i^3 \)
\( = \frac{1}{27} + i - 9 - 27i \)
\( = \left( \frac{1}{27} - 9 \right) + (i - 27i) \)
\( = \frac{1 - 243}{27} - 26i \)
Standard form: \( -\frac{242}{27} - 26i \)
Q10
\( \left( -2 - \frac{1}{3}i \right)^3 \)
Answer:
Factor out \(-1\): \( [-(2 + \frac{i}{3})]^3 = -(2 + \frac{i}{3})^3 \)
Using \( (a+b)^3 \):
\( (2)^3 + 3(2)^2(\frac{i}{3}) + 3(2)(\frac{i}{3})^2 + (\frac{i}{3})^3 \)
\( = 8 + 4i + 6(\frac{-1}{9}) + \frac{-i}{27} \)
\( = 8 + 4i - \frac{2}{3} - \frac{i}{27} \)
\( = (8 - \frac{2}{3}) + i(4 - \frac{1}{27}) \)
\( = \frac{22}{3} + i\frac{107}{27} \)
Applying the negative sign from the start:
Standard form: \( -\frac{22}{3} - i\frac{107}{27} \)
Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.
Q11
\( 4 - 3i \)
Answer:
Let \( z = 4 - 3i \). The multiplicative inverse is \( z^{-1} = \frac{1}{4-3i} \).
Multiply numerator and denominator by conjugate \( 4+3i \):
\( \frac{4+3i}{(4-3i)(4+3i)} = \frac{4+3i}{16 - (9i^2)} = \frac{4+3i}{16+9} \)
\( = \frac{4+3i}{25} \)
Inverse: \( \frac{4}{25} + i\frac{3}{25} \)
Q12
\( \sqrt{5} + 3i \)
Answer:
Let \( z = \sqrt{5} + 3i \). Inverse is \( \frac{1}{\sqrt{5}+3i} \).
\( \frac{\sqrt{5}-3i}{(\sqrt{5}+3i)(\sqrt{5}-3i)} = \frac{\sqrt{5}-3i}{5 - 9i^2} = \frac{\sqrt{5}-3i}{5+9} \)
Inverse: \( \frac{\sqrt{5}}{14} - i\frac{3}{14} \)
Q13
\( -i \)
Answer:
Inverse is \( \frac{1}{-i} \).
Multiply by \( i \):
\( \frac{i}{-i^2} = \frac{i}{-(-1)} = i \)
Inverse: \( 0 + i \)
Q14
Express the following expression in the form of \( a + ib \):
\( \frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - i\sqrt{2})} \)
\( \frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - i\sqrt{2})} \)
Answer:
Numerator: \( (3)^2 - (i\sqrt{5})^2 = 9 - (5i^2) = 9 + 5 = 14 \)
Denominator: \( \sqrt{3} + \sqrt{2}i - \sqrt{3} + \sqrt{2}i = 2\sqrt{2}i \)
Expression: \( \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i} \)
Rationalize denominator (multiply by \( i \)):
\( \frac{7i}{\sqrt{2}i^2} = \frac{7i}{-\sqrt{2}} = -\frac{7}{\sqrt{2}}i \)
Standard form: \( 0 - i\frac{7\sqrt{2}}{2} \)
Q1
Evaluate: \( \left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3 \)
Answer:
\( i^{18} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1 \)
\( \frac{1}{i} = -i \), so \( (\frac{1}{i})^{25} = (-i)^{25} = -(i^{25}) = -(i^{24} \cdot i) = -i \)
Expression becomes: \( [-1 - i]^3 = -(1+i)^3 \)
\( = -[1 + 3i + 3i^2 + i^3] \)
\( = -[1 + 3i - 3 - i] = -[-2 + 2i] \)
Result: \( 2 - 2i \)
Q2
For any two complex numbers \( z_1 \) and \( z_2 \), prove that \( \text{Re}(z_1 z_2) = \text{Re } z_1 \text{Re } z_2 - \text{Im } z_1 \text{Im } z_2 \)
Answer:
Let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \).
\( z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2) \)
\( = x_1x_2 + i x_1y_2 + i y_1x_2 + i^2 y_1y_2 \)
\( = (x_1x_2 - y_1y_2) + i(x_1y_2 + y_1x_2) \)
\( \text{Re}(z_1 z_2) = x_1x_2 - y_1y_2 \)
Since \( x_1 = \text{Re } z_1, x_2 = \text{Re } z_2, y_1 = \text{Im } z_1, y_2 = \text{Im } z_2 \):
\( \text{Re}(z_1 z_2) = \text{Re } z_1 \text{Re } z_2 - \text{Im } z_1 \text{Im } z_2 \).
Q3
Reduce \( \left( \frac{1}{1-4i} - \frac{2}{1+i} \right) \left( \frac{3-4i}{5+i} \right) \) to the standard form.
Answer:
First bracket:
\( \frac{1(1+i) - 2(1-4i)}{(1-4i)(1+i)} = \frac{1+i-2+8i}{1+i-4i-4i^2} = \frac{-1+9i}{5-3i} \)
Multiply by second part:
\( \left( \frac{-1+9i}{5-3i} \right) \left( \frac{3-4i}{5+i} \right) = \frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2} \)
\( = \frac{33+31i}{28-10i} \)
Multiply by conjugate \( 28+10i \):
\( \frac{(33+31i)(28+10i)}{28^2+10^2} = \frac{924 + 330i + 868i + 310i^2}{784+100} \)
\( = \frac{614 + 1198i}{884} = \frac{307}{442} + i\frac{599}{442} \)
Q4
If \( x - iy = \sqrt{\frac{a-ib}{c-id}} \), prove that \( (x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \)
Answer:
Given \( x - iy = \sqrt{\frac{a-ib}{c-id}} \).
Taking modulus on both sides:
\( |x - iy| = \left| \sqrt{\frac{a-ib}{c-id}} \right| = \sqrt{\left| \frac{a-ib}{c-id} \right|} \)
\( \sqrt{x^2+y^2} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}} \)
Squaring both sides:
\( x^2+y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} \)
Squaring again:
\( (x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \).
Q5
If \( z_1 = 2 - i, z_2 = 1 + i \), find \( \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right| \)
Answer:
Numerator: \( (2-i) + (1+i) + 1 = 4 \)
Denominator: \( (2-i) - (1+i) + 1 = 2 - 2i \)
Expression: \( \left| \frac{4}{2-2i} \right| = \frac{|4|}{|2-2i|} \)
\( |2-2i| = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \)
Result: \( \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \)
Q6
If \( a + ib = \frac{(x+i)^2}{2x^2+1} \), prove that \( a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \)
Answer:
Given \( a + ib = \frac{(x+i)^2}{2x^2+1} \).
Take modulus of both sides:
\( |a+ib| = \left| \frac{(x+i)^2}{2x^2+1} \right| \)
\( \sqrt{a^2+b^2} = \frac{|x+i|^2}{|2x^2+1|} \)
Since \( 2x^2+1 \) is real, its modulus is itself. \( |x+i| = \sqrt{x^2+1} \).
\( \sqrt{a^2+b^2} = \frac{(\sqrt{x^2+1})^2}{2x^2+1} = \frac{x^2+1}{2x^2+1} \)
Squaring both sides:
\( a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \)
Q7
Let \( z_1 = 2 - i, z_2 = -2 + i \). Find (i) \( \text{Re}\left(\frac{z_1 z_2}{\bar{z}_1}\right) \), (ii) \( \text{Im}\left(\frac{1}{z_1 \bar{z}_1}\right) \)
Answer:
(i) \( \text{Re}\left(\frac{z_1 z_2}{\bar{z}_1}\right) \)
\( z_1 z_2 = (2-i)(-2+i) = -4 + 2i + 2i - i^2 = -3 + 4i \)
\( \bar{z}_1 = 2+i \)
\( \frac{z_1 z_2}{\bar{z}_1} = \frac{-3+4i}{2+i} \cdot \frac{2-i}{2-i} = \frac{-6 + 3i + 8i - 4i^2}{4+1} = \frac{-2 + 11i}{5} \)
Real part = \( -\frac{2}{5} \)
(ii) \( \text{Im}\left(\frac{1}{z_1 \bar{z}_1}\right) \)
\( z_1 \bar{z}_1 = |z_1|^2 = 2^2 + (-1)^2 = 5 \)
\( \frac{1}{z_1 \bar{z}_1} = \frac{1}{5} = \frac{1}{5} + 0i \)
Imaginary part = \( 0 \)
Q8
Find the real numbers \( x \) and \( y \) if \( (x - iy)(3 + 5i) \) is the conjugate of \( -6 - 24i \).
Answer:
Conjugate of \( -6 - 24i \) is \( -6 + 24i \).
Expand LHS: \( (x - iy)(3 + 5i) = 3x + 5xi - 3yi - 5y(i^2) = (3x + 5y) + i(5x - 3y) \).
Equate Real and Imaginary parts:
- \( 3x + 5y = -6 \)
- \( 5x - 3y = 24 \)
Solving equations:
Multiply (1) by 3 and (2) by 5:
\( 9x + 15y = -18 \)
\( 25x - 15y = 120 \)
Add: \( 34x = 102 \Rightarrow x = 3 \)
Substitute \( x=3 \) in (1): \( 9 + 5y = -6 \Rightarrow 5y = -15 \Rightarrow y = -3 \)
Answer: \( x = 3, y = -3 \)
Q9
Find the modulus of \( \frac{1+i}{1-i} - \frac{1-i}{1+i} \)
Answer:
Simplify terms:
\( \frac{1+i}{1-i} = \frac{(1+i)^2}{2} = \frac{2i}{2} = i \)
\( \frac{1-i}{1+i} = \frac{(1-i)^2}{2} = \frac{-2i}{2} = -i \)
Expression: \( i - (-i) = 2i \)
Modulus: \( |2i| = 2 \)
Q10
If \( (x+iy)^3 = u+iv \), then show that \( \frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2) \)
Answer:
\( (x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 \)
\( = x^3 + 3ix^2y - 3xy^2 - iy^3 \)
\( = (x^3 - 3xy^2) + i(3x^2y - y^3) \)
Comparing with \( u+iv \):
\( u = x^3 - 3xy^2 \Rightarrow \frac{u}{x} = x^2 - 3y^2 \)
\( v = 3x^2y - y^3 \Rightarrow \frac{v}{y} = 3x^2 - y^2 \)
\( \frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2) = 4x^2 - 4y^2 = 4(x^2 - y^2) \)
Q11
If \( \alpha \) and \( \beta \) are different complex numbers with \( |\beta| = 1 \), then find \( \left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right| \)
Answer:
We need \( |w| \) where \( w = \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \). Consider \( |w|^2 = w \bar{w} \).
\( w \bar{w} = \left( \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right) \left( \frac{\bar{\beta} - \bar{\alpha}}{1 - \alpha \bar{\beta}} \right) \)
Using \( \beta \bar{\beta} = |\beta|^2 = 1 \):
Numerator: \( \beta\bar{\beta} - \beta\bar{\alpha} - \alpha\bar{\beta} + \alpha\bar{\alpha} = 1 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2 \)
Denominator: \( 1 - \alpha\bar{\beta} - \bar{\alpha}\beta + \alpha\bar{\alpha}\beta\bar{\beta} = 1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2(1) \)
Numerator = Denominator.
So \( |w|^2 = 1 \Rightarrow |w| = 1 \).
Q12
Find the number of non-zero integral solutions of the equation \( |1 - i|^x = 2^x \).
Answer:
\( |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \)
Equation becomes \( (\sqrt{2})^x = 2^x \)
\( 2^{x/2} = 2^x \)
Equating powers: \( \frac{x}{2} = x \Rightarrow x = 2x \Rightarrow x = 0 \).
The question asks for non-zero integral solutions.
Since \( x=0 \) is the only solution, there are 0 non-zero solutions.
Q13
If \( (a + ib)(c + id)(e + if)(g + ih) = A + iB \), then show that \( (a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2 \)
Answer:
Take modulus on both sides:
\( |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB| \)
Since \( |z_1 z_2| = |z_1| |z_2| \):
\( \sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2} = \sqrt{A^2+B^2} \)
Squaring both sides gives the required result.
Q14
If \( \left( \frac{1+i}{1-i} \right)^m = 1 \), then find the least positive integral value of \( m \).
Answer:
Simplify the inner term:
\( \frac{1+i}{1-i} = \frac{(1+i)^2}{1^2+1^2} = \frac{1+2i-1}{2} = i \)
Equation becomes \( i^m = 1 \).
The powers of \( i \) repeat: \( i^1=i, i^2=-1, i^3=-i, i^4=1 \).
The least positive integer \( m \) is 4.