11th NCERT Complex Numbers and Quadratic Equations Miscellaneous Exercise 5 Questions 20

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Question (1)

Evaluate: \[{\left[ {{i^{18}} + {{\left( {\frac{1}{i}} \right)}^{25}}} \right]^3}\]

Solution

\[{\left[ {{i^{18}} + {{\left( {\frac{1}{i}} \right)}^{25}}} \right]^3}\] \[ = {\left[ {{{\left( {{i^2}} \right)}^9} + {{\left[ {\frac{1}{{{i^2}}}} \right]}^{13}}i} \right]^3}\] \[ = {\left( { - 1 - i} \right)^3}\] \[ = - {\left( {1 + i} \right)^3}\] \[ = - \left[ {1 + {i^3} + 3i + 3{i^2}} \right]\] \[ = - \left[ {1 - i + 3i - 3} \right]\] \[ = - \left( {2i - 2} \right)\] \[ = 2 - 2i\]

Question (2)

For any two complex numbers z₁ and z₂, prove that Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Solution

\[{z_1} = {x_1} + i{y_1},\;\;{z_2} = {x_2} + i{y_2},\] \[{z_1} \cdot {z_2} = \left( {{x_1} + i{y_1}} \right)\left( {{x_2} + i{y_2}} \right)\] \[ = {x_1}{x_2} + {x_1}{y_2}i + {x_2}{y_1}i + {y_1}{y_2}{i^2}\] \[ = {x_1}{x_2} + {x_1}{y_2}i + {x_2}{y_1}i - {y_1}{y_2}\] \[ = {x_1}{x_2} - {y_1}{y_2} + \left( {{x_1}{y_2} + {x_2}{y_1}} \right)i\] \[{\mathop{\rm Re}\nolimits} \left( {{z_1}{z_2}} \right) = {x_1}{x_2} - {y_1}{y_2}\] ∴ Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Question (3)

Reduce to the standard form. \[\left( {\frac{1}{{1 - 4i}} - \frac{2}{{1 + i}}} \right)\left( {\frac{{3 - 4i}}{{5 + i}}} \right)\]

Solution

\[\left( {\frac{1}{{1 - 4i}} - \frac{2}{{1 + i}}} \right)\left( {\frac{{3 - 4i}}{{5 + i}}} \right)\] \[ = \left( {\frac{{1 + i - 2\left( {1 - 4i} \right)}}{{\left( {1 - 4i} \right)\left( {1 + i} \right)}}} \right)\left( {\frac{{3 - 4i}}{{5 + i}}} \right)\] \[ = \left( {\frac{{1 + i - 2 + 8i}}{{1 + i - 4i - 4{i^2}}}} \right)\left( {\frac{{3 - 4i}}{{5 + i}}} \right)\] \[ = \frac{{\left( { - 1 + 9i} \right)\left( {3 - 4i} \right)}}{{\left( {5 - 3i} \right)\left( {5 + i} \right)}}\] \[ = \frac{{ - 3 + 4i + 27i - 36{i^2}}}{{25 + 5i - 15i - 3{i^2}}}\] \[ = \frac{{33 + 31i}}{{28 - 10i}}\] \[ = \frac{{\left( {33 + 31i} \right)\left( {28 + 10i} \right)}}{{\left( {28 - 10i} \right)\left( {28 + 10i} \right)}}\] \[ = \frac{{924 + 330i + 868i + 310{i^2}}}{{784 - 100{i^2}}}\] \[ = \frac{{614 + 1198i}}{{884}}\] \[ = \frac{{2\left( {307 + 599i} \right)}}{{884}}\] \[ = \frac{{307 + 599i}}{{442}}\]

Question (4)

\[x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} ,prove\;that{\left( {{x^2} + {y^2}} \right)^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}\]

Solution

\[x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} \] \[x - iy = \sqrt {\frac{{a - ib}}{{c + id}} \times \frac{{c + id}}{{c + id}}} \] \[x - iy = \sqrt {\frac{{ac + adi - ibc - {i^2}bd}}{{{c^2} - {i^2}{d^2}}}} \] \[x - iy = \sqrt {\frac{{ac + bd + \left( {ad - bc} \right)i}}{{{c^2} - {i^2}{d^2}}}} \] \[x - iy = \sqrt {\frac{{ac + bd}}{{{c^2} + {d^2}}} + \frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}i} \].......(1) \[x + iy = \sqrt {\frac{{ac + bd}}{{{c^2} + {d^2}}} - \frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}i} \].......(2) Multiplying (1) and (2) , we get, \[\left( {x + iy} \right)\left( {x - iy} \right) = \left( {\sqrt {\frac{{ac + bd}}{{{c^2} + {d^2}}} + \frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}i} } \right)\left( {\sqrt {\frac{{ac + bd}}{{{c^2} + {d^2}}} - \frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}i} } \right)\] \[\left( {{x^2} - {i^2}{y^2}} \right) = \left( {\sqrt {{{\left( {\frac{{ac + bd}}{{{c^2} + {d^2}}}} \right)}^2} - {{\left( {\frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}} \right)}^2}{i^2}} } \right)\] \[\left( {{x^2} + {y^2}} \right) = \sqrt {{{\frac{{\left( {ac + bd} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}}^2} + {{\frac{{\left( {ad - bc} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}}^2}} \] \[\left( {{x^2} + {y^2}} \right) = \sqrt {\frac{{{a^2}{c^2} + 2acbd + {b^2}{d^2} + {a^2}{d^2} - 2adbc + {b^2}{c^2}}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}} \] \[\left( {{x^2} + {y^2}} \right) = \sqrt {\frac{{{a^2}\left( {{c^2} + {d^2}} \right) + {b^2}\left( {{d^2} + {c^2}} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}} \] \[\left( {{x^2} + {y^2}} \right) = \sqrt {\frac{{\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}} \] \[\left( {{x^2} + {y^2}} \right) = \sqrt {\frac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {{c^2} + {d^2}} \right)}}} \] Squaring on both sides we get, \[{\left( {{x^2} + {y^2}} \right)^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}\]

Question (5)

Convert the following in the polar form: (i)\[\frac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\] (ii) \[\frac{{1 + 3i}}{{\left( {1 - 2i} \right)}}\]

Solution

(i) \[\frac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\] \[ = \frac{{1 + 7i}}{{4 - 4i + {i^2}}}\] \[ = \frac{{1 + 7i}}{{3 - 4i}}\] \[ = \frac{{1 + 7i}}{{3 - 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}\] \[ = \frac{{3 + 4i + 21i + 28{i^2}}}{{9 - 16{i^2}}}\] \[ = \frac{{ - 25 + 25i}}{{25}}\] \[ = - 1 + i\] Z = -1 + i
Comparing to standard form we get, a = -1 , b = 1
\[r = \sqrt {{a^2} + {b^2}} \] \[ = \sqrt {1 + 1} = \sqrt 2 \] \[a = r\cos \theta ,\quad b = r\sin \theta \] \[-1 = \sqrt 2 \cos \theta ,\quad 1 = \sqrt 2 \sin \theta \] \[\cos \theta = \frac{-1}{{\sqrt 2 }},\quad \sin \theta = \frac{1}{{\sqrt 2 }}\] So θ lies in 2^nd quadrant .
\[\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\] So polar form of complex number is given by \[r\left( {\cos \theta + i\sin \theta } \right)\] \[ = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\] (ii) \[\frac{{1 + 3i}}{{\left( {1 - 2i} \right)}}\] \[\frac{{1 + 3i}}{{\left( {1 - 2i} \right)}} \times \frac{{1 + 2i}}{{1 + 2i}}\] \[ = \frac{{1 + 2i + 3i + 6{i^2}}}{{1 - 4{i^2}}}\] \[ = \frac{{ - 5 + 5i}}{5}\] \[ = - 1 + i\] Z = -1 + i
Comparing to standard form we get, a = -1 , b = 1
\[r = \sqrt {{a^2} + {b^2}} \] \[ = \sqrt {1 + 1} = \sqrt 2 \] \[a = r\cos \theta ,\quad b = r\sin \theta \] \[-1 = \sqrt 2 \cos \theta ,\quad 1 = \sqrt 2 \sin \theta \] \[\cos \theta = \frac{-1}{{\sqrt 2 }},\quad \sin \theta = \frac{1}{{\sqrt 2 }}\] So θ lies in 2^nd quadrant .
\[\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\] So polar form of complex number is given by \[r\left( {\cos \theta + i\sin \theta } \right)\] \[ = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\]

Question (6)

Solve the equation\[3{x^2} - 4x + \frac{{20}}{3} = 0\]

Solution

\[3{x^2} - 4x + \frac{{20}}{3} = 0\] \[9{x^2} - 12x + 20 = 0\] Comparing to standard form we get, a = 9, b = -12, c = 20.
\[D = {b^2} - 4ac\] \[ = {\left( { - 12} \right)^2} - 4\left( 9 \right)\left( {20} \right)\] \[ = 144 - 720\] \[ = - 576 = 576{i^2}\] \[D = \ 24i\] \[x = \frac{{ - b \pm \sqrt D }}{{2a}}\] \[ = \frac{{12 \pm 24i}}{{2\left( 9 \right)}}\] \[ = \frac{{6\left( {2 \pm 4i} \right)}}{{18}}\] \[ = \frac{{2 \pm 4i}}{3}\]

Question (7)

Solve the equation \[{x^2} - 2x + \frac{3}{2} = 0\]

Solution

\[{x^2} - 2x + \frac{3}{2} = 0\] \[2{x^2} - 4x + 3 = 0\] Comparing to standard form we get, a = 2, b = -4, c = 3.
\[D = {b^2} - 4ac\] \[ = {\left( { - 4} \right)^2} - 4\left( 2 \right)\left( 3 \right)\] \[ = 16 - 24\] \[ = - 8 = 8{i^2}\] \[D = 2\sqrt 2 i\] \[x = \frac{{ - b \pm \sqrt D }}{{2a}}\] \[ = \frac{{4 \pm 2\sqrt 2 i}}{{2\left( 2 \right)}}\] \[ = \frac{{2\left( {2 \pm \sqrt 2 i} \right)}}{4}\] \[ = \frac{{2 \pm \sqrt 2 i}}{2}\]

Question (8)

Solve the equation 27x² – 10x + 1 = 0

Solution

27x² – 10x + 1 = 0
Comparing to standard form we get, a = 27, b = -10, c = 1.
\[D = {b^2} - 4ac\] \[ = {\left( { - 10} \right)^2} - 4\left( 27 \right)\left( 1 \right)\] \[ = 100 - 108\] \[ = -8 = 8{i^2}\] \[D = \sqrt {8} i = 2\sqrt {2} i\] \[x = \frac{{ - b \pm \sqrt D }}{{2a}}\] \[ = \frac{{10 \pm 2\sqrt {2} i}}{{2\left( {27} \right)}}\] \[ = \frac{{2\left( {5 \pm \sqrt {2} i} \right)}}{{2\left( {27} \right)}}\] \[ = \frac{{5 \pm \sqrt {2} i}}{{27}}\]

Question (9)

Solve the equation 21x² – 28x + 10 = 0

Solution

21x² – 28x + 10 = 0
Comparing to standard form we get, a = 21, b = -28, c = 10.
\[D = {b^2} - 4ac\] \[ = {\left( { - 28} \right)^2} - 4\left( 21 \right)\left( 10 \right)\] \[ =784 - 840\] \[ = - 56 = 56{i^2}\] \[D = \sqrt {56} i = 2\sqrt {14} i\] \[x = \frac{{ - b \pm \sqrt D }}{{2a}}\] \[ = \frac{{28 \pm 2\sqrt {14} i}}{{2\left( {21} \right)}}\] \[ = \frac{{2\left( {14 \pm \sqrt {14} i} \right)}}{{2\left( {21} \right)}}\] \[ = \frac{{14 \pm \sqrt {14} i}}{{21}}\]

Question (10)

If z₁ = 2 - i , z₂ = 1 + i , find \[\left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right|\].

Solution

\[\left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right|\] \[ = \left| {\frac{{2 - i + 1 + i + 1}}{{2 - i - 1 - i + 1}}} \right|\] \[ = \left| {\frac{4}{{2 - 2i}}} \right|\] \[ = \frac{{\left| 4 \right|}}{{\left| {2 - 2i} \right|}}\] \[ = \frac{4}{{\sqrt {4 + 4} }}\] \[ = \frac{4}{{2\sqrt 2 }}\] \[ = \sqrt 2 \]

Question (11)

If\[a + ib = \frac{{{{\left( {x + i} \right)}^2}}}{{2{x^2} + 1}}\] prove that \[{a^2} + {b^2} = \frac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\]

Solution

\[a + ib = \frac{{{{\left( {x + i} \right)}^2}}}{{2{x^2} + 1}}\] \[ = \frac{{{x^2} + 2xi + {i^2}}}{{2{x^2} + 1}}\] \[ = \frac{{{x^2} - 1 + 2xi}}{{2{x^2} + 1}}\] \[ = \frac{{{x^2} - 1}}{{2{x^2} + 1}} + \frac{{2x}}{{2{x^2} + 1}}i.....(1)\] \[a - ib = \frac{{{x^2} - 1}}{{2{x^2} + 1}} - \frac{{2x}}{{2{x^2} + 1}}i.....(2)\] Multiply (1) and (2), \[{a^2} - {i^2}{b^2} = {\left( {\frac{{{x^2} - 1}}{{2{x^2} + 1}}} \right)^2} - {\left( {\frac{{2x}}{{2{x^2} + 1}}i} \right)^2}\] \[{a^2} + {b^2} = \frac{{{{\left( {{x^2} - 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}} - \frac{{4{x^2}{i^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\] \[ = \frac{{{x^4} - 2{x^2} + 1 + 4{x^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\] \[ = \frac{{{x^4} + 2{x^2} + 1}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\] \[ = \frac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\] Hence proved.

Question (12)

Let z₁ = 2 - i , z₂ = -2 + i. Find (i)\[{\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }}} \right)\] , (ii) \[{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{{Z_1}\overline {{Z_1}} }}} \right)\]

Solution

(i)\[{\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }}} \right)\] , \[{z_1} = 2 - i,\overline {{z_1}} = 2 + i\] \[\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }} = \frac{{\left( {2 - i} \right)\left( { - 2 + i} \right)}}{{2 + i}}\] \[ = \frac{{ - 4 + 2i + 2i - {i^2}}}{{2 + i}}\] \[ = \frac{{ - 3 + 4i}}{{2 + i}}\] \[ = \frac{{\left( { - 3 + 4i} \right)\left( {2 - i} \right)}}{{\left( {2 + i} \right)\left( {2 - i} \right)}}\] \[ = \frac{{ - 6 + 3i + 8i - 4{i^2}}}{{4 - {i^2}}}\] \[ = \frac{{ - 2 + 11i}}{5}\] \[ = - \frac{2}{5} + \frac{{11}}{5}i\] \[{\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }}} \right) = - \frac{2}{5}\] (ii) \[{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{{Z_1}\overline {{Z_1}} }}} \right)\] \[{z_1} = 2 - i,\overline {{z_1}} = 2 + i\] \[\frac{1}{{{Z_1}\overline {{Z_1}} }} = \frac{1}{{(2 - i)(2 + i)}}\] \[ = \frac{1}{{4 - {i^2}}}\] \[ = \frac{1}{5}\] \[{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{{Z_1}\overline {{Z_1}} }}} \right) = 0\]

Question (13)

Find the modulus and argument of the complex number\[\frac{{1 + 2i}}{{1 - 3i}}\]

Solution

\[\frac{{1 + 2i}}{{1 - 3i}}\] \[ = \frac{{\left( {1 + 2i} \right)\left( {1 + 3i} \right)}}{{\left( {1 - 3i} \right)\left( {1 + 3i} \right)}}\] \[ = \frac{{1 + 3i + 2i + 6{i^2}}}{{1 - 9{i^2}}}\] \[ = \frac{{ - 5 + 5i}}{{10}}\] \[ = - \frac{1}{2} + \frac{1}{2}i\] Comparing to standard form we get, a = - 1/2, and b = 1/2
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \] \[ = \sqrt {{{\left( {\frac{{ - 1}}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} \] \[ = \sqrt {\frac{1}{4} + \frac{1}{4}} \] \[ = \frac{1}{{\sqrt 2 }}\] \[a = r\cos \theta \;\quad b = r\sin \theta \] \[ - \frac{1}{2} = \frac{1}{{\sqrt 2 }}\cos \theta ,\quad \frac{1}{2} = \frac{1}{{\sqrt 2 }}\sin \theta \] \[\cos \theta = - \frac{1}{{\sqrt 2 }},\quad \sin \theta = \frac{1}{{\sqrt 2 }}\] θ lies in 2nd quadrant. \[\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\] So argumet θ = 3π/4.

Question (14)

Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Solution

The conjugate of –6 – 24i. is -6 + 24i
(x – iy) (3 + 5i)= -6 + 24i
3x + 5xi - 3yi - 5yi² = - 6 + 24i
( 3x + 5y) + (5x - 3y)i = -6 + 24i
⇒ 3x + 5y = -6 .. (1) and 5x - 3y = 24 .... (2)
Multiply (1) by 3 and (2) by 5 we get,
9x + 15y = -18 .....(3)
25x - 15y = 120 .... (4)
Add (3) and (4) we get, 34x = 102
∴ x = 3
Replacing this value of x in (1) we get,
9 + 5y = -6
5y = -15
y = -3.

Question (15)

Find the modulus of \[\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}\].

Solution

\[\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}\] \[ = \frac{{{{\left( {1 + i} \right)}^2} - {{\left( {1 - i} \right)}^2}}}{{1 - {i^2}}}\] \[ = \frac{{1 + 2i + {i^2} - 1 + 2i - {i^2}}}{2}\] \[ = \frac{{4i}}{2} = 2i\] Comparind to standard form we get, a = 0 , b = 2 \[\left| z \right| = \sqrt {{a^2} + {b^2}} \] \[ = \sqrt {0 + 4} \] \[ = 2\]

Question (16)

If (x + iy)³ = u + iv, then show that \[\frac{u}{x} + \frac{v}{y} = 4\left( {{x^2} - {y^2}} \right)\]

Solution

\[u + iv = {\left( {x + iy} \right)^3}\] \[ = {x^3} + {i^3}{y^3} + 3xyi\left( {x + yi} \right)\] \[ = {x^3} - {y^3}i + 3{x^2}yi + 3x{y^2}{i^2}\] \[ = {x^3} - 3x{y^2} + i\left( {3{x^2}y - {y^3}} \right)\] \[ \Rightarrow u = {x^3} - 3x{y^2},v = 3{x^2}y - {y^3}\] \[\frac{u}{x} + \frac{v}{y} = \frac{{{x^3} - 3x{y^2}}}{x} + \frac{{3{x^2}y - {y^3}}}{y}\] \[ = {x^2} - 3{y^2} + 3{x^2} - {y^2}\] \[ = 4{x^2} - 4{y^2}\] \[ = 4\left( {{x^2} - {y^2}} \right)\]

Question (17)

If α and β are different complex numbers with | β | = 1, then find \[\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|\].

Solution

Let α = a + bi nd β = x + yi
|β | = 1 \[\alpha = a + bi,\;\overline \alpha = a - bi\] \[\left| \alpha \right| = \left| {\overline \alpha } \right|\] \[\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}\] \[\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right| = \frac{{\left| {\beta - \alpha } \right|}}{{\left| {1 - \overline \alpha \beta } \right|}}\] \[ = \frac{{\left| \beta \right| - \left| \alpha \right|}}{{1 - \left| {\overline \alpha \left| \beta \right|} \right|}}\] \[ = \frac{{1 - \left| \alpha \right|}}{{1 - \left| \alpha \right|(1)}}\] \[ = 1\]

Question (18)

Find the number of non-zero integral solutions of the equation.\[{\left| {1 - i} \right|^x} = {2^x}\]

Solution

\[{\left| {1 - i} \right|^x} = {2^x}\] \[{\left( {\sqrt {{1^2} + {{( - 1)}^2}} } \right)^x} = {2^x}\] \[{\left( {\sqrt 2 } \right)^x} = {2^x}\] \[{2^{\frac{x}{2}}} = {2^x}\] \[ \Rightarrow \frac{x}{2} = x\] \[ \Rightarrow x = 2x\] \[ \Rightarrow - x = 0\] \[ \Rightarrow x = 0\]

Question (19)

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Solution

We will use the property \[\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|\] \[\left( {a + ib} \right)\left( {c + id} \right)\left( {e + if} \right)\left( {g + ih} \right) = A + iB\] \[\left| {a + ib} \right|\left| {c + id} \right|\left| {e + if} \right|\left| {g + ih} \right| = \left| {A + iB} \right|\] \[\left( {\sqrt {{a^2} + {b^2}} } \right)\left( {\sqrt {{c^2} + {d^2}} } \right)\left( {\sqrt {{e^2} + {f^2}} } \right)\left( {\sqrt {{g^2} + {h^2}} } \right) = \sqrt {{A^2} + {B^2}} \] Squaring on both sides, \[\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\left( {{e^2} + {f^2}} \right)\left( {{g^2} + {h^2}} \right) = {A^2} + {B^2}\] Hence proved.

Question (20)

If \[{\left( {\frac{{1 + i}}{{1 - i}}} \right)^m} = 1\], then find the least positive integral value of m.

Solution

\[{\left( {\frac{{1 + i}}{{1 - i}}} \right)^m} = 1\] \[{\left( {\frac{{1 + i}}{{1 - i}} \times \frac{{1 + i}}{{1 + i}}} \right)^m} = 1\] \[\left( {\frac{{1 + 2i + {i^2}}}{{1 - {i^2}}}} \right) = 1\] \[{\left( {\frac{{1 + 2i - 1}}{2}} \right)^m} = 1\] \[{i^m} = 1\] \[{i^m} = {i^{4k}}\] \[m = 4k\] m = 4k , k > 0, k ∈ N.
So m is multiple of 4. least value of m = 4. as least value of k = 1.