Miscellaneous Exercise Complex Numbers and Quadratic-equations Solution - NCERT Class 11

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Chapter 5 Ex.5.ME Question 1

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

Solution

\[\begin{align}  {{\left[ {{i}^{18}}+{{\left( \frac{1}{i} \right)}^{25}} \right]}^{3}}&={{\left[ {{i}^{4\times 4+2}}+\frac{1}{{{i}^{4\times 6+1}}} \right]}^{3}} \\ & ={{\left[ {{\left( {{i}^{4}} \right)}^{4}}\times {{i}^{2}}+\frac{1}{{{\left( {{i}^{4}} \right)}^{6}}\times i} \right]}^{3}} \\ & ={{\left[ {{i}^{2}}+\frac{1}{i} \right]}^{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{4}}=1 \right] \\ & ={{\left[ -1+\frac{1}{i}\times \frac{i}{i} \right]}^{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\ & ={{\left[ -1+\frac{i}{{{i}^{2}}} \right]}^{3}} \\ & ={{\left[ -1-i \right]}^{3}} \\ & ={{\left( -1 \right)}^{3}}{{\left[ 1+i \right]}^{3}} \\ & =-\left[ {{1}^{3}}+{{i}^{3}}+3\times 1\times i\left( 1+i \right) \right] \\ & =-\left[ 1+{{i}^{3}}+3i+3{{i}^{2}} \right] \\ & =-\left[ 1-i+3i-3 \right] \\ & =-\left[ -2+2i \right] \\ & =2-2i \end{align}\]

Chapter 5 Ex.5.ME Question 2

For any two complex numbers \({z_1}\)and \({z_2}\), prove that

\[{\mathop{\rm Re}\nolimits} \;\left( {{z_1}{z_2}} \right) = {\mathop{\rm Re}\nolimits} {z_1}{\mathop{\rm Re}\nolimits} {z_2} - {\mathop{\rm Im}\nolimits} {z_1}{\mathop{\rm Im}\nolimits} {z_2}\]

Solution

Let \({z_1} = {x_1} + i{y_1}\;{\rm{and}}\;{z_2} = {x_2} + i{y_2}\)

\[\begin{align}  {{z}_{1}}{{z}_{2}}&=\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right) \\ & ={{x}_{1}}\left( {{x}_{2}}+i{{y}_{2}} \right)+i{{y}_{1}}\left( {{x}_{2}}+i{{y}_{2}} \right) \\ & ={{x}_{1}}{{x}_{2}}+i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}+{{i}^{2}}{{y}_{1}}{{y}_{2}} \\ & ={{x}_{1}}{{x}_{2}}+i{{x}_{1}}{{y}_{2}}+i{{y}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because {{i}^{2}}=-1 \right] \\ & =\left( {{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}} \right)+i\left( {{x}_{1}}{{y}_{2}}+{{y}_{1}}{{x}_{2}} \right) \end{align}\]

\[\begin{align} \Rightarrow {\mathop{\rm Re}\nolimits} \;\left( {{z_1}{z_2}} \right) &= {x_1}{x_2} - {y_1}{y_2}\\ \Rightarrow {\mathop{\rm Re}\nolimits} \;\left( {{z_1}{z_2}} \right)& = {\mathop{\rm Re}\nolimits} {z_1}{\mathop{\rm Re}\nolimits} {z_2} - {\mathop{\rm Im}\nolimits} {z_1}{\mathop{\rm Im}\nolimits} {z_2}\end{align}\]

Hence, proved.

Chapter 5 Ex.5.ME Question 3

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

Solution

\[\begin{align}\left( {\frac{1}{{1 - 4i}} - \frac{2}{{1 + i}}} \right)\left( {\frac{{3 - 4i}}{{5 + i}}} \right) &= \left[ {\frac{{\left( {1 + i} \right) - 2\left( {1 - 4i} \right)}}{{\left( {1 - 4i} \right)\left( {1 + i} \right)}}} \right]\left[ {\frac{{3 - 4i}}{{5 + i}}} \right]\\ &= \left[ {\frac{{1 + i - 2 + 8i}}{{1 + i - 4i - 4{i^2}}}} \right]\left[ {\frac{{3 - 4i}}{{5 + i}}} \right]\\ &= \left[ {\frac{{ - 1 + 9i}}{{5 - 3i}}} \right]\left[ {\frac{{3 - 4i}}{{5 + i}}} \right]\\ &= \left[ {\frac{{ - 3 + 4i + 27i - 36{i^2}}}{{25 + 5i - 15i - 3{i^2}}}} \right]\\ &= \frac{{33 + 31i}}{{28 - 10i}}\\ &= \frac{{33 + 31i}}{{2\left( {14 - 5i} \right)}}\\ &= \frac{{\left( {33 + 31i} \right)}}{{2\left( {14 - 5i} \right)}} \times \frac{{\left( {14 + 5i} \right)}}{{\left( {14 + 5i} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \begin{array}{l}{\text{On multiplying numerator and}}\\{\rm{denominator by }}\left( {14 + 5i} \right)\end{array} \right]\\ &= \frac{{462 + 165i + 434i + 155{i^2}}}{{2\left[ {{{\left( {14} \right)}^2} - {{\left( {5i} \right)}^2}} \right]}}\\ &= \frac{{307 + 599i}}{{2\left( {196 - 25{i^2}} \right)}}\\ &= \frac{{307 + 599i}}{{2\left( {221} \right)}}\\ &= \frac{{307}}{{442}} + \frac{{599}}{{442}}i\end{align}\]

This is the required standard form.

Chapter 5 Ex.5.ME Question 4

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

Solution

\[\begin{align}x - iy &= \sqrt {\frac{{a - ib}}{{c - id}}} = \sqrt {\frac{{a - ib}}{{c - id}} \times \frac{{c + id}}{{c + id}}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \begin{array}{l}{\text{On multiplying numerator and}}\\{\rm{denominator by}}\left( {c + id} \right)\end{array} \right]\\ &= \sqrt {\frac{{\left( {ac + bd} \right) + i\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}} \\{\left( {x - iy} \right)^2} &= \frac{{\left( {ac + bd} \right) + i\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}\\{x^2} - {y^2} - 2ixy &= \frac{{\left( {ac + bd} \right)}}{{{c^2} + {d^2}}} + i\frac{{\left( {ad - bc} \right)}}{{{c^2} + {d^2}}}\end{align}\]

On comparing real and imaginary parts, we obtain

\[{x^2} - {y^2} = \frac{{ac + bd}}{{{c^2} + {d^2}}},\; - 2xy = \frac{{ad - bc}}{{{c^2} + {d^2}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

Since,

\[\begin{align}{\left( {{x^2} + {y^2}} \right)^2} &= {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}\\ &= {\left( {\frac{{ac + bd}}{{{c^2} + {d^2}}}} \right)^2} + {\left( {\frac{{ad - bc}}{{{c^2} + {d^2}}}} \right)^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using}}\;\left( 1 \right)} \right]\\ &= \frac{{{a^2}{c^2} + {b^2}{d^2} + 2acbd + {a^2}{d^2} + {b^2}{c^2} - 2abcd}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}\\ &= \frac{{{a^2}{c^2} + {b^2}{d^2} + {a^2}{d^2} + {b^2}{c^2}}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}\\& = \frac{{{a^2}\left( {{c^2} + {d^2}} \right) + {b^2}\left( {{c^2} + {d^2}} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}\\ &= \frac{{\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)}}{{{{\left( {{c^2} + {d^2}} \right)}^2}}}\\ &= \frac{{\left( {{a^2} + {b^2}} \right)}}{{\left( {{c^2} + {d^2}} \right)}}\end{align}\]

Hence, proved.

Chapter 5 Ex.5.ME Question 5

Convert the following in the polar form:

(i) \(\frac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\)

(ii) \(\frac{{1 + 3i}}{{1 - 2i}}\)

Solution

(i) Here,

\[\begin{align}z &= \frac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}\\ &= \frac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}} = \frac{{1 + 7i}}{{4 + {i^2} - 4i}} = \frac{{1 + 7i}}{{4 - 1 - 4i}}\\ &= \frac{{1 + 7i}}{{3 - 4i}} \times \frac{{3 + 4i}}{{3 + 4i}} = \frac{{3 + 4i + 21i + 28{i^2}}}{{{3^2} + {4^2}}}\\ &= \frac{{3 + 25i - 28}}{{25}} = \frac{{ - 25 + 25i}}{{25}}\\ &= - 1 + i\end{align}\]

Let \(r\cos \theta = - 1\) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\[\begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta &= {\left( { - 1} \right)^2} + {1^2}\\{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) &= 1 + 1\\{r^2} &= 2\\r &= \sqrt 2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},{\rm{ }}r > 0} \right]\end{align}\]

Therefore,

\(\sqrt 2 \cos \theta = - 1\) and \(\sqrt 2 \sin \theta = 1\)

\( \Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}\) and \(\sin \theta = \frac{1}{{\sqrt 2 }}\)

Since, \(\theta \) lies in the quadrant II, \(\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\)

Hence,

\[\begin{align}z &= r\cos \theta + ir\sin \theta \\& = \sqrt 2 \cos \frac{{3\pi }}{4} + i\sqrt 2 \sin \frac{{3\pi }}{4}\\ &= \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\end{align}\]

This is the required polar form.

(ii) Here,

\[\begin{align}z &= \frac{{1 + 3i}}{{1 - 2i}}\\ &= \frac{{1 + 3i}}{{1 - 2i}} \times \frac{{1 + 2i}}{{1 + 2i}}\\ &= \frac{{1 + 2i + 3i + 6{i^2}}}{{{1^2} + {2^2}}} = \frac{{1 + 5i - 6}}{5}\\ &= \frac{{ - 5 + 5i}}{5} = - 1 + i\end{align}\]

Let \(r\cos \theta = - 1\) and \(r\sin \theta = 1\)

On squaring and adding, we obtain

\[\begin{align}{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta &= {\left( { - 1} \right)^2} + {1^2}\\{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) &= 1 + 1\\{r^2} &= 2\\r &= \sqrt 2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},\;r > 0} \right]\end{align}\]

Therefore,

\(\sqrt 2 \cos \theta = - 1\) and \(\sqrt 2 \sin \theta = 1\)

\( \Rightarrow \cos \theta = - \frac{1}{{\sqrt 2 }}\) and \(\sin \theta = \frac{1}{{\sqrt 2 }}\)

Since, \(\theta \) lies in the quadrant II, \(\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\)

Hence,

\[\begin{array}{c}z &= r\cos \theta + ir\sin \theta \\ &= \sqrt 2 \cos \frac{{3\pi }}{4} + i\sqrt 2 \sin \frac{{3\pi }}{4}\\ &= \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)\end{array}\]

This is the required polar form.

Chapter 5 Ex.5.ME Question 6

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

Solution

The given quadratic equation is \(3{x^2} - 4x + \frac{{20}}{3} = 0\)

This equation can also be written as \(9{x^2} - 12x + 20 = 0\)

On comparing this equation with \(a{x^2} + bx + c = 0\), we obtain \(a = 9,\;b = - 12\) and \(c = 20\)

Therefore, the discriminant of the given equation is

\[\begin{align}D &= {b^2} - 4ac\\ &= {\left( { - 12} \right)^2} - 4 \times 9 \times 20\\ &= 144 - 720\\& = - 576\end{align}\]

Hence, the required solutions are

\[\begin{align}  \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -12 \right)\pm \sqrt{-576}}{2\times 9} \\ & =\frac{12\pm \sqrt{576}i}{18}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{12\pm 24i}{18} \\ & =\frac{6\left( 2\pm 4i \right)}{18} \\ & =\frac{2\pm 4i}{3} \\ & =\frac{2}{3}\pm \frac{4}{3}i \end{align}\]

Chapter 5 Ex.5.ME Question 7

Solve the equation \({x^2} - 2x + \frac{3}{2} = 0\)

Solution

The given quadratic equation is \({x^2} - 2x + \frac{3}{2} = 0\)

This equation can also be written as \(2{x^2} - 4x + 3 = 0\)

On comparing this equation with \(a{x^2} + bx + c = 0\), we obtain

\(a = 2,\;b = - 4\) and \(c = 3\)

Therefore, the discriminant of the given equation is

\[\begin{align}D &= {b^2} - 4ac\\ &= {\left( { - 4} \right)^2} - 4 \times 2 \times 3\\& = 16 - 24\\& = - 8\end{align}\]

Hence, the required solutions are

\[\begin{align}  \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -4 \right)\pm \sqrt{-8}}{2\times 2} \\ & =\frac{4\pm 2\sqrt{2}i}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{2\pm \sqrt{2}i}{2} \\ & =1\pm \frac{\sqrt{2}}{2}i \end{align}\]

Chapter 5 Ex.5.ME Question 8

Solve the equation \(27{x^2} - 10x + 1 = 0\)

Solution

The given quadratic equation is \(27{x^2} - 10x + 1 = 0\)

On comparing this equation with \(a{x^2} + bx + c = 0\), we obtain

\(a = 27,\;b = - 10\) and \(c = 1\)

Therefore, the discriminant of the given equation is

\[\begin{align}D &= {b^2} - 4ac\\& = {\left( { - 10} \right)^2} - 4 \times 27 \times 1\\ &= 100 - 108\\ &= - 8\end{align}\]

Hence, the required solutions are

\[\begin{align}  \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -10 \right)\pm \sqrt{-8}}{2\times 27} \\ & =\frac{10\pm 2\sqrt{2}i}{54}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{5\pm \sqrt{2}i}{27} \\ & =\frac{5}{27}\pm \frac{\sqrt{2}}{27}i \end{align}\]

Chapter 5 Ex.5.ME Question 9

Solve the equation \(21{x^2} - 28x + 10 = 0\)

Solution

The given quadratic equation is \(21{{x}^{2}}-28x+10=0\)

On comparing this equation with\(a{{x}^{2}}+bx+c=0\), we obtain

\(a = 21,\;b = - 28\) and \(c = 10\)

Therefore, the discriminant of the given equation is

\[\begin{align}D &= {b^2} - 4ac\\ &= {\left( { - 28} \right)^2} - 4 \times 21 \times 10\\ &= 784 - 840\\ &= - 56\end{align}\]

Hence, the required solutions are

\[\begin{align}  \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -28 \right)\pm \sqrt{-56}}{2\times 21} \\ & =\frac{28\pm \sqrt{56}i}{42}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{28\pm 2\sqrt{14}i}{42} \\ & =\frac{28}{42}\pm \frac{2\sqrt{14}}{42}i \\ & =\frac{2}{3}\pm \frac{\sqrt{14}}{21}i \end{align}\]

Chapter 5 Ex.5.ME Question 10

If \({z_1} = 2 - i,\;{z_2} = 1 + i,\)find \(\left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right|\)

Solution

\[{z_1} = 2 - i,\;{z_2} = 1 + i\]

Therefore,

\[\begin{align}\left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right| &= \left| {\frac{{\left( {2 - i} \right) + \left( {1 + i} \right) + 1}}{{\left( {2 - i} \right) - \left( {1 + i} \right) + 1}}} \right|\\ &= \left| {\frac{4}{{2 - 2i}}} \right| = \left| {\frac{4}{{2\left( {1 - i} \right)}}} \right|\\ &= \left| {\frac{2}{{1 - i}} \times \frac{{1 + i}}{{1 + i}}} \right| = \left| {\frac{{2\left( {1 + i} \right)}}{{\left( {{1^2} - {i^2}} \right)}}} \right|\\ &= \left| {\frac{{2\left( {1 + i} \right)}}{{1 + 1}}} \right|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{i^2} = - 1} \right]\\ &= \left| {\frac{{2\left( {1 + i} \right)}}{2}} \right|\\ &= \left| {1 + i} \right| = \sqrt {{1^2} + {1^2}} \\ &= \sqrt 2 \end{align}\]

Thus, the value of \(\left| {\frac{{{z_1} + {z_2} + 1}}{{{z_1} - {z_2} + 1}}} \right|\)is \(\sqrt 2 \).

Chapter 5 Ex.5.ME 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

\(a + ib = \frac{{{{\left( {x + i} \right)}^2}}}{{2{x^2} + 1}}\)

\[\begin{align}a + ib = \frac{{{{\left( {x + i} \right)}^2}}}{{2{x^2} + 1}} &= \frac{{{x^2} + {i^2} + i2x}}{{2{x^2} + 1}}\\ &= \frac{{{x^2} - 1 + i2x}}{{2{x^2} + 1}}\\& = \frac{{{x^2} - 1}}{{2{x^2} + 1}} + i\left( {\frac{{2x}}{{2{x^2} + 1}}} \right)\end{align}\]

On comparing real and imaginary parts, we obtain

\(a = \frac{{{x^2} - 1}}{{2{x^2} + 1}}\) and \(b = \frac{{2x}}{{2{x^2} + 1}}\)

Since,

\[\begin{align}{a^2} + {b^2} &= {\left( {\frac{{{x^2} - 1}}{{2{x^2} + 1}}} \right)^2} + {\left( {\frac{{2x}}{{2{x^2} + 1}}} \right)^2}\\ &= \frac{{{x^4} + 1 - 2{x^2} + 4{x^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\\ &= \frac{{{x^4} + 1 + 2{x^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\\{a^2} + {b^2} &= \frac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}\end{align}\]

Hence, proved.

Chapter 5 Ex.5.ME Question 12

Let \({z_1} = 2 - i,\;{z_2} = - 2 + i\). Find

(i) \({\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{{{\bar z}_1}}}} \right)\),

(ii) \({\mathop{\rm Im}\nolimits} \left( {\frac{1}{{{z_1}{{\bar z}_1}}}} \right)\)

 

Solution

\[{z_1} = 2 - i,\;{z_2} = - 2 + i\]

(i) It is given that

\[\begin{align}{z_1}{z_2} &= \left( {2 - i} \right)\left( { - 2 + i} \right)\\ &= - 4 + 2i + 2i - {i^2}\\ &= - 4 + 4i - \left( { - 1} \right)\\ &= - 3 + 4i\end{align}\]

Now, \({\bar z_1} = 2 + i\)

Hence,\(\frac{{{z_1}{z_2}}}{{{{\bar z}_1}}} = \frac{{ - 3 + 4i}}{{2 + i}}\)

On multiplying numerator and denominator by \(\left( {2 - i} \right)\), we obtain

\[\begin{align}\frac{{{z_1}{z_2}}}{{{{\bar z}_1}}} &= \frac{{\left( { - 3 + 4i} \right)\left( {2 - i} \right)}}{{\left( {2 + i} \right)\left( {2 - i} \right)}} = \frac{{ - 6 + 3i + 8i - 4{i^2}}}{{{2^2} + {1^2}}} = \frac{{ - 6 + 11i - 4\left( { - 1} \right)}}{5}\\ &= \frac{{ - 2 + 11i}}{5} = \frac{{ - 2}}{5} + \frac{{11}}{5}i\end{align}\]

On comparing real parts, we obtain

\[{\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}{z_2}}}{{{{\bar z}_1}}}} \right) = - \frac{2}{5}\]

(ii) \(\frac{1}{{{z_1}{{\bar z}_1}}} = \frac{1}{{\left( {2 + i} \right)\left( {2 - i} \right)}} = \frac{1}{{{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}}} = \frac{1}{5}\)

On comparing imaginary parts, we obtain

\[{\mathop{\rm Im}\nolimits} \left( {\frac{1}{{{z_1}{{\bar z}_1}}}} \right) = 0\]

Chapter 5 Ex.5.ME Question 13

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

Solution

Let \(z=\frac{1+2i}{1-3i}\),

Then,

\[\begin{align}z &= \frac{{1 + 2i}}{{1 - 3i}} \times \frac{{1 + 3i}}{{1 + 3i}} = \frac{{1 + 2i + 3i + 6{i^2}}}{{{1^2} + {3^2}}} = \frac{{1 + 5i + 6\left( { - 1} \right)}}{{10}}\\& = \frac{{ - 5 + 5i}}{{10}} = \frac{{ - 5}}{{10}} + \frac{5}{{10}}i = \frac{{ - 1}}{2} + \frac{1}{2}i\end{align}\]

Let \(z = r\cos \theta + ir\sin \theta \)

i.e., \(r\cos \theta = - \frac{1}{2}\) and \(r\sin \theta = \frac{1}{2}\)

On squaring and adding, we obtain

\[\begin{align}\;\;\;\;{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) &= {\left( { - \frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^2}\\ \Rightarrow {r^2} &= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\\ \Rightarrow r &= \frac{1}{{\sqrt 2 }}\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Conventionally}},r > 0} \right]\end{align}\]

Therefore,

\(\frac{1}{{\sqrt 2 }}\cos \theta  =  - \frac{1}{2}\) and \(\frac{1}{\sqrt{2}}\sin \theta =\frac{1}{2}\)

\(\Rightarrow \cos \theta =\frac{-1}{\sqrt{2}}\) and \(\frac{1}{{\sqrt 2 }}\cos \theta = - \frac{1}{2}\) \(\sin \theta =\frac{1}{\sqrt{2}}\)

Since, \(\theta \) lies in the quadrant II, \(\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\)

Thus, the modulus and argument of the given complex number are \(\frac{1}{{\sqrt 2 }}\) and \(\frac{3\pi }{4}\) respectively.

Chapter 5 Ex.5.ME Question 14

Find the real numbers \(x\) and \(y\)if \(\left( x-iy \right)\left( 3+5i \right)\) is the conjugate of \(-6-24i\)

Solution

Let,

\[\begin{align}z &= \left( {x - iy} \right)\left( {3 + 5i} \right)\\ &= 3x - i3y + i5x - {i^2}5y\\ &= 3x + i5x - i3y + 5y\\ &= \left( {3x + 5y} \right) + i\left( {5x - 3y} \right)\end{align}\]

Then, \(\bar z = \left( {3x + 5y} \right) - i\left( {5x - 3y} \right)\)

It is given that, \(\bar z = - 6 - 24i\)

Therefore, \(\left( {3x + 5y} \right) - i\left( {5x - 3y} \right) = - 6 - 24i\)

Equating real and imaginary parts, we obtain

\[\begin{align}3x + 5y = - 6\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\5x - 3y = 24\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

Multiplying equation (\(1\)) by \(3\) and equation (\(2\)) by \(5\) and then adding them, we obtain

\[\begin{align}9x + 15y = - 18\\25x - 15y = 120\end{align}\]

On adding both equations we get,

\[\begin{align}34x&= 102\\x&= \frac{{102}}{{34}}\\x&= 3\end{align}\]

Putting the value of \(x\) in equation (\(1\)), we obtain

\[\begin{align}3\left( 3 \right) + 5y &= - 6\\5y &= - 6 - 9\\y &= \frac{{ - 15}}{5}\\y &= - 3\end{align}\]

Thus, the values of \(x = 3\) and \(y = - 3\)

Chapter 5 Ex.5.ME Question 15

Find the modulus of \(\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}\)

Solution

\[\begin{align}\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}&= \frac{{{{\left( {1 + i} \right)}^2} - {{\left( {1 - i} \right)}^2}}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}\\ &= \frac{{1 + {i^2} + 2i - 1 - {i^2} + 2i}}{2}\\ &= \frac{{4i}}{2}\\ &= 2i\end{align}\]

Therefore,

\[\begin{align}\left| {\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}} \right| &= \left| {2i} \right|\\& = \sqrt {{2^2}} \\ &= 2\end{align}\]

Chapter 5 Ex.5.ME Question 16

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

Solution

It is given that \({\left( {x + iy} \right)^3} = u + iv\)

\[\begin{align}& \Rightarrow {x^3} + {\left( {iy} \right)^3} + 3 \times x \times iy\left( {x + iy} \right) = u + iv\\ &\Rightarrow {x^3} + {i^3}{y^3} + 3{x^2}yi + 3x{y^2}{i^2} = u + iv\\ &\Rightarrow {x^3} - i{y^3} + 3{x^2}yi - 3x{y^2} = u + iv\\ &\Rightarrow \left( {{x^3} - 3x{y^2}} \right) + i\left( {3{x^2}y - {y^3}} \right) = u + iv\end{align}\]

On equating real and imaginary parts, we obtain

\(u = {x^3} - 3x{y^2},\;v = 3{x^2}y - {y^3}\)

Therefore,

\[\begin{align}\frac{u}{x} + \frac{v}{y} &= \frac{{{x^3} - 3x{y^2}}}{x} + \frac{{3{x^2}y - {y^3}}}{y}\\
& = \frac{{x\left( {{x^2} - 3{y^2}} \right)}}{x} + \frac{{y\left( {3{x^2} - {y^2}} \right)}}{y}\\ &= {x^2} - 3{y^2} + 3{x^2} - {y^2}\\ &= 4{x^2} - 4{y^2}\\ &= 4\left( {{x^2} - {y^2}} \right)\end{align}\]

Hence, \(\frac{u}{x} + \frac{v}{y} = 4\left( {{x^2} - {y^2}} \right)\) proved.

Chapter 5 Ex.5.ME Question 17

If \(\alpha \) and are different complex numbers with \(\left| \beta \right| = 1,\)then find \(\left| {\frac{{\beta - \alpha }}{{1 - \bar \alpha \beta }}} \right|\)

Solution

Let \(\alpha = a + ib\) and \(\beta = x + iy\)

It is given that, \(\left| \beta \right| = 1\)

Therefore,

\[\begin{align}\sqrt {{x^2} + {y^2}} &= 1\\{x^2} + {y^2} &= 1\;\;\;\;\;\;\;\;\;\; \ldots \left( i \right)\end{align}\]

\[\begin{align}  \left| \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right|&=\left| \frac{\left( x+iy \right)-\left( a+ib \right)}{1-\left( a-ib \right)\left( x+iy \right)} \right| \\ & =\left| \frac{\left( x-a \right)+i\left( y-b \right)}{1-\left( ax+iay-ibx+by \right)} \right| \\ & =\left| \frac{\left( x-a \right)+i\left( y-b \right)}{\left( 1-ax-by \right)+i\left( bx-ay \right)} \right| \\ & =\frac{\left| \left( x-a \right)+i\left( y-b \right) \right|}{\left| \left( 1-ax-by \right)+i\left( bx-ay \right) \right|}\ \ \ \ \ \ \ \ \ \ \ \left[ \because \left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=\frac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|} \right] \end{align}\]

\[\begin{align}\left| {\frac{{\beta - \alpha }}{{1 - \bar \alpha \beta }}} \right| &= \frac{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}} }}{{\sqrt {{{\left( {1 - ax - by} \right)}^2} + {{\left( {bx - ay} \right)}^2}} }}\\ &= \frac{{\sqrt {{x^2} + {a^2} - 2ax + {y^2} + {b^2} - 2by} }}{{\sqrt {1 + {a^2}{x^2} + {b^2}{y^2} - 2ax + 2abxy - 2by + {b^2}{x^2} + {a^2}{y^2} - 2abxy} }}\\ &= \frac{{\sqrt {\left( {{x^2} + {y^2}} \right) + {a^2} + {b^2} - 2ax - 2by} }}{{\sqrt {1 + {a^2}\left( {{x^2} + {y^2}} \right) + {b^2}\left( {{y^2} + {x^2}} \right) - 2ax - 2by} }}\\ &= \frac{{\sqrt {1 + {a^2} + {b^2} - 2ax - 2by} }}{{\sqrt {1 + {a^2} + {b^2} - 2ax - 2by} }}\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( i \right)} \right]\\ &= 1\end{align}\]

Thus, \(\left| {\frac{{\beta - \alpha }}{{1 - \bar \alpha \beta }}} \right| = 1\)

Chapter 5 Ex.5.ME Question 18

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

Solution

\[\begin{align}{\left| {1 - i} \right|^x} &= {2^x}\\\left( {\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \right) &= {2^x}\\{\left( {\sqrt 2 } \right)^x} &= {2^x}\\{2^{x/2}} &= {2^x}\\\frac{x}{2} &= x\\x &= 2x\\2x - x &= 0\\x &= 0\end{align}\]

Thus, \(0\) is the only integral solution of the given equation.

Therefore, the number of non-zero integral solutions of the given equation is \(0\).

Chapter 5 Ex.5.ME Question 19

If \(\left( {a + ib} \right)\left( {c + id} \right)\left( {e + if} \right)\left( {g + ih} \right) = A + iB,\) then show that:

   \(\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)={{A}^{2}}+{{B}^{2}}\)

Solution

It is given that, \(\left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right)=A+iB\)

Therefore,

\[\begin{align}  \left| \left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right) \right|&=\left| A+iB \right| \\  \left| \left( a+ib \right) \right|\times \left| \left( c+id \right) \right|\times \left| \left( e+if \right) \right|\times \left| \left( g+ih \right) \right|&=\left| A+iB \right|\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \left| {{z}_{1}}{{z}_{2}} \right|=\left| {{z}_{1}} \right|\left| {{z}_{2}} \right| \right] \\  \sqrt{{{a}^{2}}+{{b}^{2}}}\times \sqrt{{{c}^{2}}+{{d}^{2}}}\times \sqrt{{{e}^{2}}+{{f}^{2}}}\times \sqrt{{{g}^{2}}+{{h}^{2}}}&=\sqrt{{{A}^{2}}+{{B}^{2}}} \\  \sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)}&=\sqrt{{{A}^{2}}+{{B}^{2}}} \\ \end{align}\]

On squaring both sides, we obtain

\[\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)={{A}^{2}}+{{B}^{2}}\]

Hence, proved.

Chapter 5 Ex.5.ME Question 20

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

Solution

It is given that \({\left( {\frac{{1 + i}}{{1 - i}}} \right)^m} = 1\)

\[\begin{align}{\left( {\frac{{1 + i}}{{1 - i}} \times \frac{{1 + i}}{{1 + i}}} \right)^m} &= 1\\{\left( {\frac{{{{\left( {1 + i} \right)}^2}}}{{{1^2} + {1^2}}}} \right)^m} &= 1\\{\left( {\frac{{{1^2} + {i^2} + 2i}}{2}} \right)^m} &= 1\\{\left( {\frac{{1 - 1 + 2i}}{2}} \right)^m} &= 1\\{\left( {\frac{{2i}}{2}} \right)^m} A&= 1\\{i^m} &= 1\\{i^m} &= {i^{4k}}\end{align}\]

Hence, \(m = 4k,\) where \(k\) is some integer.

Since, the least positive integer is \(1\), \(m = 4 \times 1 = 4\)

Thus, the least positive integral value of \(m = 4\)