Complex Number Exercise 6.2 | Basic Mathematics Solution [NEB UPDATED]

Exercise 6.2

1) Express each of the following complex number in the form of a + ib.

a) (2 + 5i)

Solution:

(2 + 5i) + (1 + i) = 2 + 5i + 1 – i = 3 + 4i

Which is in the form of a + ib where, a = 3, b = 4. 

b) (2 + 5i) – (4 – i)

Solution:

(2 + 5i) – (4 – i) = 2 + 5i – 4 + I = - 2 + 6i

Which is in the form of a + ib where, a = -2, b = 6. 

c) (2 + 3i)(3 – 2i)

Solution:

(2 + 3i)(3 – 2i) = 6 – 4i + 9i – 6i² = 6 + 5i + 6 = 12 + 5i

Which is in the form of a + ib where, a = 12, b = 5 

d) $\frac{{3 + 4{\rm{i}}}}{{4 - 3{\rm{i}}}}$

Solution:

$\frac{{3 + 4{\rm{i}}}}{{4 - 3{\rm{i}}}}$ = $\frac{{3 + 4{\rm{i}}}}{{4 - 3{\rm{i}}}}{\rm{*}}\frac{{4 + 3{\rm{i}}}}{{4 + 3{\rm{i}}}}$ = $\frac{{12 + 9{\rm{i}} + 16{\rm{i}} + 12{{\rm{i}}^2}}}{{{4^2} - \left( {3{{\rm{i}}^2}} \right)}}$

= $\frac{{12 + 25{\rm{i}} - 12}}{{16 - 9{{\rm{i}}^2}}}$ = $\frac{{25{\rm{i}}}}{{16 + 9}}$ = $\frac{{25{\rm{i}}}}{{25}}$ = i = 0 + i.

Which is in the form of a + ib where, a = 0, b = 1 

e) $\frac{{\rm{i}}}{{2 + {\rm{i}}}}{\rm{\: }}$

Solution:

$\frac{{\rm{i}}}{{2 + {\rm{i}}}}{\rm{\: }}$= $\frac{{\rm{i}}}{{2 + {\rm{i}}}}{\rm{*}}\frac{{2 - {\rm{i}}}}{{2 - {\rm{i}}}}$ = $\frac{{2{\rm{i}} - {{\rm{i}}^2}}}{{{2^2} - {1^2}}}$ = $\frac{{2{\rm{i}} - \left( { - 1} \right)}}{{4 - \left( { - 1} \right)}}$ = $\frac{{2{\rm{i}} + 1}}{{4 + 1}}$ = $\frac{{2{\rm{i}}}}{5}$ + $\frac{1}{5}$ = $\frac{1}{5}$ + $\frac{2}{5}$i

Which is in the form of a + ib where, a = $\frac{1}{5}$, b = $\frac{2}{5}$. 

f) $\frac{{1 - {\rm{i}}}}{{{{\left( {1 + {\rm{i}}} \right)}^2}}}$

Solution:

$\frac{{1 - {\rm{i}}}}{{{{\left( {1 + {\rm{i}}} \right)}^2}}}$ = $\frac{{1 - {\rm{i}}}}{{1 + 2{\rm{i}} + {{\rm{i}}^2}}}$ = $\frac{{1 - {\rm{i}}}}{{1 + 2{\rm{i}} - 1}}$ = $\frac{{1 - {\rm{i}}}}{{2{\rm{i}}}}$ = $\frac{{1 - {\rm{i}}}}{{2{\rm{i}}}}$ * $\frac{{2{\rm{i}}}}{{2{\rm{i}}}}$

= $\frac{{2{{\rm{i}}^2} - 2{{\rm{i}}^2}}}{{4{{\rm{i}}^2}}}$ = $\frac{{2{\rm{i}} - 2\left( { - 1} \right)}}{{4\left( { - 1} \right)}}$ = $\frac{{\left( {2{\rm{i}} + 2} \right)}}{{ - 4}}$ = $\frac{2}{{ - 4}} - \frac{{2{\rm{i}}}}{4}$ = $ - \frac{1}{2} - \frac{1}{2}{\rm{i}}$

Which is in the form of a + ib where, a = $ - \frac{1}{2}$ , b = $ - \frac{1}{2}$. 

g) $\frac{{2 - \sqrt { - 25} }}{{1 - \sqrt { - 16} }}$

Solution:

$\frac{{2 - \sqrt { - 25} }}{{1 - \sqrt { - 16} }}$ = $\frac{{2 - \sqrt { - 1{\rm{*}}25} }}{{1 - \sqrt {{{\rm{i}}^2}{\rm{*}}{4^2}} }}$ = $\frac{{2 - 5{\rm{i}}}}{{1 - 4{\rm{i}}}}{\rm{*}}\frac{{1 + 4{\rm{i}}}}{{1 + 4{\rm{i}}}}$

= $\frac{{2 + 8{\rm{i}} - 5{\rm{i}} - 20{{\rm{i}}^2}}}{{{1^2} - 16{{\rm{i}}^2}}}$ = $\frac{{2 + 3{\rm{i}} - 20\left( { - 1} \right)}}{{1 + 16}}$ = $\frac{{2 + 3{\rm{i}} + 20}}{{17}}$

= $\frac{{22}}{{17}}$ + $\frac{{3{\rm{i}}}}{{17}}$ Which is in the form of a + ib where, a =$\frac{{22}}{{17}}$ and b = $\frac{3}{{17}}$. 

h) $\sqrt {\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}} $ 

Solution:

$\sqrt {\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}} $ = $\sqrt {\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}{\rm{*}}\frac{{1 + {\rm{i}}}}{{1 + {\rm{i}}}}} $ = $\frac{{1 + {\rm{i}}}}{{\sqrt {1 - {{\rm{i}}^2}} }}$ = $1 + {\rm{i}}/(\sqrt {1 + 1} {\rm{\: }}$ = $\frac{1}{{\sqrt 2 }}$ + $\frac{1}{{\sqrt 2 }}$i

Which is in the form of a + ib where, a = $\frac{1}{{\sqrt 2 }}$ , b = $\frac{1}{{\sqrt 2 }}$.

 

2) If z = 2 + 3i, and w=3-2i, find ${{\rm{\bar z}}^2}$ + ${{\rm{\bar w}}^2}$.

Solution:

If z = 2 + 3i, ${\rm{\bar z}}$ = 2 – 3i, If w = 3 – 2i, ${\rm{\bar w}}$ = 3 + 2i

Now, ${{\rm{\bar z}}^2}$ + ${{\rm{\bar w}}^2}$= (2 – 3i)² + (3 + 2i)²

= 4 – 12i + 9i² + 9 + 12i + 4i²

= 4 – 12i – 9 + 9 + 12i – 4 = 0 

3) Compute the absolute values of the following:

a) 1+ 2i

Solution:

Absolute value of 1 + 2i = |1 + 2i| = $\sqrt {{1^2} + {2^2}} $ = $\sqrt 5 $. 

b) 1 + $\sqrt 3 $i

Solution:

Absolute value of 1 + $\sqrt 3 $i = |1 + $\sqrt 3 $i| = $\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} $ = 2.

c)(1 + 2i)(2 + i) 

Solution:

Absolute value of (1 + 2i)(2 + i) = |(1 + 2i)(2 + i)| = |1 + 2i||2 + i| =  $\sqrt {1 + 4} $.$\sqrt {4 + 1} $ = 5 

d) (3 + 4i)(3 – 4i)

Solution:

Absolute value of (3 + 4i)(3 – 4i) = |(3 + 4i)(3 – 4i)|= |3 + 4i||3 + 4i| = $\sqrt {{3^2} + {4^2}} $ = $\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} $ = $\sqrt {25} .\sqrt {25} $ = 25. 

e)  (1 + i)^-1

Solution:

Absolute value of (1 + i)^-1 = |(1 + i)^-1| = $\left| {\frac{1}{{1 + {\rm{i}}}}} \right|$.

= $\left| {\frac{1}{{1 + {\rm{i}}}}{\rm{*}}\frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}}} \right|$ = $\left| {\frac{{1 - {\rm{i}}}}{{{1^2} - {{\rm{i}}^2}}}} \right|$ = $\left| {\frac{{1 - {\rm{i}}}}{{1 + 1}}} \right|$

= $\left| {\frac{1}{2} - \frac{{\rm{i}}}{2}} \right|$ = $\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} $ = $\sqrt {\frac{1}{4} + \frac{1}{4}} $ = $\sqrt {\frac{2}{4}} $ = $\frac{1}{{\sqrt 2 }}$

f) $\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}$

Solution:

Absolute value of $\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}$ = $\left| {\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}} \right|$ = $\frac{{\left| {1 + {\rm{i}}} \right|}}{{\left| {1 - {\rm{i}}} \right|}}$ = $\frac{{\sqrt {{1^2} + {1^2}} }}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} }}$ = $\frac{{\sqrt {1 + 1} }}{{\sqrt {1 + 1} }}$ = $\frac{{\sqrt 2 }}{{\sqrt 2 }}$ = 1.

4) If z= 1+2i and w=2-i, verify that: 

a)$\overline {{\rm{zw}}} $ = ${\rm{\bar z}}$.${\rm{\bar w}}$

Solution:

If z = 1 + 2i, ${\rm{\bar z}}$ = 1 – 2i and if w = 2 – i, ${\rm{\bar w}}$ = 2 + i

zw = (1 + 2i)(2 – i ) = 2 – i + 4i – 2i² = 2 + 3i + 2 = 4 + 3i.

Then, $\overline {{\rm{zw}}} $ = 4 – 3i

Again, ${\rm{\bar z}}$.${\rm{\bar w}}$ = (1 – 2i)(2 + i) 2 + i – 4i – 2i² = 2 – 3i + 2 = 4 – 4i.

Hence, $\overline {{\rm{zw}}} $ = ${\rm{\bar z}}$.${\rm{\bar w}}$.

b)$\overline {\left( {\frac{{\rm{z}}}{{\rm{w}}}} \right)} $ = $\frac{{{\rm{\bar z}}}}{{{\rm{\bar w}}}}$

Solution:

= $\left( {\frac{{\rm{z}}}{{\rm{w}}}} \right)$ = $\frac{{1 + 2{\rm{i}}}}{{2 - {\rm{i}}}}$ = $\frac{{1 + 2{\rm{i}}}}{{2 - {\rm{i}}}}{\rm{*}}\frac{{2 + {\rm{i}}}}{{2 + {\rm{i}}}}$ = $\frac{{2 + {\rm{i}} + 4{\rm{i}} + 2{{\rm{i}}^2}}}{{4 - {{\rm{i}}^2}}}$ = $\frac{{2 + 5{\rm{i}} - 2}}{{4 + 1}}$ = $\frac{{5{\rm{i}}}}{5}$ = i = 0 + i.

Then, $\overline {\left( {\frac{{\rm{z}}}{{\rm{w}}}} \right)} $ = 0 – i.

Again, $\frac{{{\rm{\bar z}}}}{{{\rm{\bar w}}}}$ = $\frac{{1 - 2{\rm{i}}}}{{2 + {\rm{i}}}}$ = $\frac{{1 - 2{\rm{i}}}}{{2 + {\rm{i}}}}{\rm{*}}\frac{{2 - {\rm{i}}}}{{2 - {\rm{i}}}}$ = $\frac{{2 - {\rm{i}} - 4{\rm{i}} + 2{{\rm{i}}^2}}}{{4 - {{\rm{i}}^2}}}$

= $\frac{{2 - 5{\rm{i}} - 2}}{{4 + 1}}$ = $ - \frac{{5{\rm{i}}}}{5}$ = -i = 0 – i.

Hence, $\overline {\left( {\frac{{\rm{z}}}{{\rm{w}}}} \right)} $ = $\frac{{{\rm{\bar z}}}}{{{\rm{\bar w}}}}$. 

c) |zw| = |z|.|w|

Solution:

|zw| = |(1 + 2i)(2 – i)| = |2 – i + 4i – 2i²| = |2 + 3i + 2|

= |4 + 3i| = $\sqrt {{4^2} + {3^2}} $ = $\sqrt {16 + 9{\rm{\: }}} $ = 5.

Again, $\left| {\rm{z}} \right|$.$\left| {\rm{w}} \right|$ = $\left| {1 + 2{\rm{i}}} \right|.\left| {2 - {\rm{i}}} \right|$ = $\sqrt {{1^2} + {2^2}} .\sqrt {{2^2} + {{\left( { - 1} \right)}^2}} $ = $\sqrt {1 + 4} $.$\sqrt {4 + 1} $ = $\sqrt 5 $.$\sqrt 5 $ = 5.

Hence, |zw| = |z|.|w| 

d)  |z + w| < |z| + |w|

Solution:

|z + w| = |1 + 2i + 2 – i| = |3 + i| = $\sqrt {{3^2} + {1^2}} $ = $\sqrt {10} $.

Again, |z| +|w| = |1 + 2i| + |2 – i|= $\sqrt {1 + 4}  + \sqrt {4 + 1} $ = $\sqrt 5 $ + $\sqrt 5 $ = 2$\sqrt 5 $ = $\sqrt {{2^2}{\rm{*}}5} $ = $\sqrt {20} $

Hence, |z + w| < |z| + |w|.

5) Prove that $\frac{{{\rm{\bar z}}}}{{{{\left( {\left| {\rm{z}} \right|} \right)}^2}}}$ is the multplicative inverse of z.

Solution:

Proof:

= $\frac{{{\rm{\bar z}}}}{{{{\left( {\left| {\rm{z}} \right|} \right)}^2}}}$.z = $\frac{{{\rm{z}}.{\rm{\bar z}}}}{{{{\left| {\rm{z}} \right|}^2}}}$ = $\frac{{{{\left| {\rm{z}} \right|}^2}}}{{{{\left| {\rm{z}} \right|}^2}}}$ = 1     [z.${\rm{\bar z}}$ = |z|²]

Hence, $\frac{{{\rm{\bar z}}}}{{{{\left( {\left| {\rm{z}} \right|} \right)}^2}}}$ is multiplicative inverse of z.

 

6.a) If (3 – 4i)(x + iy) = 3$\sqrt 5 $, show that 5x² + 5y² = 9 

Solution:

(3 – 4i)(x + iy) = 3$\sqrt 5 $

Taking modulus on both sides,

|(3 – 4i)(x + iy)| = |3$\sqrt 5 $|

= |3 – 4i||x + iy| = 3$\sqrt 5 $

= $\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} $.$\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $3\sqrt 5 $.

= $\sqrt {25} $.$\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = 3$\sqrt 5 $.

Squaring both sides,

Or, 25(x² + y²) = 9 * 5

So, 5x² + 5y² = 9 

b) If x + iy =$\frac{{{\rm{a}} - {\rm{ib}}}}{{{\rm{a}} + {\rm{ib}}}}{\rm{*}}\frac{{{\rm{a}} - {\rm{ib}}}}{{{\rm{a}} - {\rm{ib}}}}$, show that x² + y² = 1

Solution:

x + iy = $\frac{{{\rm{a}} - {\rm{ib}}}}{{{\rm{a}} + {\rm{ib}}}}{\rm{*}}\frac{{{\rm{a}} - {\rm{ib}}}}{{{\rm{a}} - {\rm{ib}}}}$ = $\frac{{{{\left( {{\rm{a}} - {\rm{ib}}} \right)}^2}}}{{{{\rm{a}}^2} - {{\rm{i}}^2}{{\rm{b}}^2}}}$= $\frac{{{{\rm{a}}^2} - 2{\rm{aib}} + {{\rm{i}}^2}{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ = $\frac{{\left( {{{\rm{a}}^2} - {{\rm{b}}^2}} \right) - 2{\rm{abi}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ = $\left( {\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}} \right) - \left( {\frac{{2{\rm{ab}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}} \right)$i

Comparing real and imaginary parts.

x = $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ and y = $ - \left( {\frac{{2{\rm{ab}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}} \right)$

Now, x² + y² = ${\left( {\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}} \right)^2} + {\left( {\frac{{ - 2{\rm{ab}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}} \right)^2}$ = $\frac{{{{\left( {{{\rm{a}}^2} - {{\rm{b}}^2}} \right)}^2} + 4{{\rm{a}}^2}{{\rm{b}}^2}}}{{{{\left( {{{\rm{a}}^2} + {{\rm{b}}^2}} \right)}^2}}}$ = $\frac{{{{\left( {{{\rm{a}}^2} + {{\rm{b}}^2}} \right)}^2}}}{{{{\left( {{{\rm{a}}^2} + {{\rm{b}}^2}} \right)}^2}}}$ = 1.

Hence, x² + y² = 1 

c) If $\frac{{1 - ix}}{{1 + ix}} = a - ib,prove{\rm{ }}that{\rm{ }}{{\rm{a}}^{\rm{2}}}{\rm{  +  }}{{\rm{b}}^{\rm{2}}}{\rm{  =  1}}{\rm{.}}$

Solution:

$\frac{{1 - {\rm{ix}}}}{{1 + {\rm{ix}}}}{\rm{*}}\frac{{1 - {\rm{ix}}}}{{1 - {\rm{ix}}}}$ = a – ib.

Or, $\frac{{1 - 2{\rm{ix}} + {{\rm{i}}^2}{{\rm{x}}^2}}}{{{1^2} - {{\rm{i}}^2}{{\rm{x}}^2}}}$ = a – ib.

Or, $\frac{{1 - 2{\rm{ix}} - {{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}$ = a – ib

Or, $\frac{{1 - {{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}} - \frac{{2{\rm{xi}}}}{{1 + {{\rm{x}}^2}}}$ = a – ib

Comparing real and imaginary parts.

Or, $\frac{{1 - {{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}$ = a and $\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}$ = b.

Now, a² + b² = ${\left( {\frac{{1 - {{\rm{x}}^2}}}{{1 + {{\rm{x}}^2}}}} \right)^2} + {\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)^2}$ = $\frac{{{{\left( {1 - {{\rm{x}}^2}} \right)}^2} + 4{{\rm{x}}^2}}}{{{{\left( {1 + {{\rm{x}}^2}} \right)}^2}}}$

= $\frac{{{{\left( {1 + {{\rm{x}}^2}} \right)}^2}}}{{{{\left( {1 + {{\rm{x}}^2}} \right)}^2}}}$ = 1.

Hence, a² + b² = 1. 

d) $If{\rm{ x  -  iy  =  }}\sqrt {\frac{{1 - i}}{{1 + i}}} ,prove\;{\rm{that }}{{\rm{x}}^{\rm{2}}}{\rm{  +   }}{{\rm{y}}^{\rm{2}}}{\rm{  = 1}}{\rm{.}}$

Solution:

x – iy = $\sqrt {\frac{{1 - {\rm{i}}}}{{1 + {\rm{i}}}}{\rm{*}}\frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}}} $ = $\sqrt {\frac{{{{\left( {1 - {\rm{i}}} \right)}^2}}}{{{1^2} - {{\rm{i}}^2}}}} $ = $\frac{{1 - {\rm{i}}}}{{\sqrt {1 + 1} }}$ = $\frac{1}{{\sqrt 2 }} - \frac{{\rm{i}}}{{\sqrt 2 }}$.

Comparing real and imaginary parts.

x = $\frac{1}{{\sqrt 2 }}$, y = $\frac{1}{{\sqrt 2 }}$.

Now, x² + y² = ${\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$+ ${\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$ = $\frac{1}{2} + \frac{1}{2}$ = 1.

Hence, x² + y² = 1

 

7) If z and w are two complex number, prove that |z + w|² = |z|² + |w|² + 2Re (z.${\rm{\bar w}}$)

Solution:

Let z = a + ib and w = c + id

Then z + w = a + ib + c + id = (a + c) + i(b + d)

Also, |z|² = a² + b² and |w|² = c² + d²

Or, z.${\rm{\bar w}}$ = (a + ib)(c – id) = ac – iad + ibc – i²bd.

= (ac + bd) – i(ad – bc)

Re.(z.${\rm{\bar w}}$) = (ac + bd)

Now, |z + w|² = |a + ib + c + id|² = |(a + c) + i(b + d)|²

= (a + c)² + (b + d)² = a² + 2ac + c² + b² + 2bd + d²

= (a² + b²) + (c² + d²) + 2(ac + bd)

= |z|² + |w|² + 2.Re  (z.${\rm{\bar w}}$)

Hence, |z + w|² = |z|² + |w|² + 2Re (z.${\rm{\bar w}}$)

 

8) Find the multiplicative inverse of the following complex number: 

a) (3 + i)²

Solution:

Multiplicative inverse of (3 + i)² = $\frac{1}{{{{\left( {3 + {\rm{i}}} \right)}^2}}}$.

= $\frac{1}{{9 + 6{\rm{i}} + {{\rm{i}}^2}}}$ = $\frac{1}{{9 + 6{\rm{i}} - 1}}$ =  $\frac{1}{{8 + 6{\rm{i}}}}$

= $\frac{1}{{8 + 6{\rm{i}}}}{\rm{*}}\frac{{8 - 6{\rm{i}}}}{{8 - 6{\rm{i}}}}$ = $\frac{{8 - 6{\rm{i}}}}{{64 - 36{{\rm{i}}^2}}}$ = $\frac{{8 - 6{\rm{i}}}}{{64 + 36}}$

= $\frac{{8 - 6{\rm{i}}}}{{100}}$ = $\frac{8}{{100}} - \frac{6}{{100}}$I = $\frac{2}{{25}} - \frac{3}{{50}}$i.

b) $\frac{{2 - 5{\rm{i}}}}{{6 + {\rm{i}}}}$

Solution:

Multiplicative inverse of $\frac{{2 - 5{\rm{i}}}}{{6 + {\rm{i}}}}$ = $\frac{{6 + {\rm{i}}}}{{2 - 5{\rm{i}}}}$

= $\frac{{6 + {\rm{i}}}}{{2 - 5{\rm{i}}}}{\rm{*}}\frac{{2 + 5{\rm{i}}}}{{2 + 5{\rm{i}}}}$ = $\frac{{12 + 30{\rm{i}} + 2{\rm{i}} + 5{{\rm{i}}^2}}}{{4 - 25{{\rm{i}}^2}}}$

= $\frac{{12 + 32{\rm{i}} - 5}}{{4 + 25}}$ = $\frac{{7 + 32{\rm{i}}}}{{29}}$ = $\frac{7}{{29}} + \frac{3}{{29}}$i.

 

9) If z and w  are two complex number, prove that |z| - |w| ≤ |z – w|.

Solution:

Proof:

Or, |z| = |z – w + w| ≤ |z – w| + |w|

So, |z| - |w| ≤ |z – w|

 

10) Determine the square root of the following numbers: 

a) 5 + 12i

Solution:

Let x + iy be the square root of 5 + 12i.

So that, (x + iy)² = 5 + 12i

Or, x² – y² + 2xy.i = 5 + 12i

So, x² – y² = 5 …(i)

2xy = 12 …(ii)

We have, (x² + y²)² = (x² – y²)² + 4x²y² = 5² + 12² = 169.

So, x² + y² = 13….(iii)

Solving (i) and (iii),

X² – y² = 5

X² + y² = 13

${\rm{\: }}$2x² = 18.

So, x = ± 3 then y = ± 2.

Since, xy > 0.
So, x = 3., y =2.

Or, x = -3, y = -2

Hence, the square roots are 3 + 2i and – 3 – 2i.

i.e. ±(3 + 2i).

 

b) -5 + 12i

Solution:

Let x + iy be the square root of – 5 + 12i so that (x + iy)² = - 5 + 12i

Or, x² – y² + 2xy.i = - 5 + 12i.

Equation real and imaginary parts,

Or, x² – y²  = - 5…(i)

Or, 2xy = 12 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 25 + 144 = 169.

Or, x² + y² = 13 …(iii)

Adding (i) and (iii), 2x² = 8 → x = ± 2.

Substituting the value of x in (iii), y² = 9 → y = ±3.

Since, 2xy = 12 > 0.

So, x = 2, y = 3 and x = - 2 , y = -3.

So, the required roots are 2 + 3i and – 2 – 3i.

i.e. ±(2 + 3i).

 

c) 8 + 6i

Solution:

Let x + iy be the square root of 8 + 6i so that (x + iy)² = 8 + 6i

Or, x² – y² + 2xy.i = 8 + 6i.

Equation real and imaginary parts,

Or, x² – y²  = 8…(i)

Or, 2xy = 6 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 64 + 36 = 100.

Or, x² + y² = 10 …(iii)

Adding (i) and (iii), x² = 9 → x = ± 3.

Substituting the value of x in (iii), y² = 1 → y = ±1.

Since, 2xy = 6 > 0.

So, x = 3, y = 1 and x = - 3 , y = -1.

So, the required roots are 3 + i and – 3 – i.

i.e. ±(3 + i). 

 

d) - 8 + 6i

Solution:

Let x + iy be the square root of 8 + 6i so that (x + iy)² = - 8 + 6i

Or, x² – y² + 2xy.i = - 8 + 6i.

Equation real and imaginary parts,

Or, x² – y²  = - 8…(i)

Or, 2xy = 6 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 64 + 36 = 100.

Or, x² + y² = 10 …(iii)

Adding (i) and (iii), x² = 1 → x = ± 1.

Substituting the value of x in (iii), y² = 9→ y = ±3.

Since, 2xy = 6 > 0.

So, x = 1, y = 3 and x = - 1 , y = -3.

So, the required roots are 1 + 3i and – 1 – 3i.

i.e. ±(1 + 3i).

 

e) 7 – 24i

Solution:

Let x + iy be the square root of 7 – 24i so that (x + iy)² = 7 – 24i

Or, x² – y² + 2xy.i = 7 – 24i

Equation real and imaginary parts,

Or, x² – y²  = 7…(i)

Or, 2xy = -24 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 49 + 576 = 625

Or, x² + y² = 25 …(iii)

Adding (i) and (iii), x² = 16 à x = ± 4.

Substituting the value of x in (iii), y² = 9 → y = ±3.

Since, 2xy = - 24 < 0.

So, x = 4, y = - 3 → 4 – 3i and x = - 4, y = 3→ - 4 + 3i.

So, the required roots are i.e. ±(4 – 3i).

 

f) - 7 + 24i

Solution:

Let x + iy be the square root of  - 7 + 24i so that (x + iy)² =  - 7 + 24i

Or, x² – y² + 2xy.i = - 7 + 24i

Equation real and imaginary parts,

Or, x² – y²  = - 7…(i)

Or, 2xy = 24 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 49 + 576 = 625

Or, x² + y² = 25 …(iii)

Adding (i) and (iii), x² = 9 → x = ± 3.

Substituting the value of x in (iii), y² = 16à y = ±4.

Since, 2xy = 25 > 0.

So, x = 3, y = 4→3 + 4i and x = - 3, y = - 4à - 3 – 4i

So, the required roots are i.e. ±(3 + 4i).

 

g) i

Solution:

Let x + iy be the square root of  - 7 + 24i so that (x + iy)² = i = 0 + i.

Or, x² – y² + 2xy.i = 0 + i

Equation real and imaginary parts,

Or, x² – y²  = 0…(i)

Or, 2xy = 1 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 0 + 1 = 1.

Or, x² + y² = 1 …(iii)

Adding (i) and (iii), x² = $\frac{1}{2}$→ x = ± $\frac{1}{{\sqrt 2 }}$.

Substituting the value of x in (iii), y² = $\frac{1}{2}$→ y = ± $\frac{1}{{\sqrt 2 }}$.

Since, 2xy = 1 > 0.

So, x = $\frac{1}{{\sqrt 2 }}$, y = $\frac{1}{{\sqrt 2 }}$à$\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}$i = $\frac{1}{{\sqrt 2 }}$(1 + i) and x = - $\frac{1}{{\sqrt 2 }}$, y = - $\frac{1}{{\sqrt 2 }}$à$ - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}$I = $\frac{1}{{\sqrt 2 }}$(- 1 – i).

So, the required roots are i.e. ± $\frac{1}{{\sqrt 2 }}$ (1 + i).

 

h) 12 – 5i

Solution:

Let x + iy be the square root of  12 – 5i so that (x + iy)² = 12 – 5i

Or, x² – y² + 2xy.i = 12 – 5i

Equation real and imaginary parts,

Or, x² – y²  = 12…(i)

Or, 2xy = -5 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 144 + 25 = 169.

Or, x² + y² = 13 …(iii)

Adding (i) and (iii), x² = $\frac{{25}}{2}$→ x = ± $\frac{5}{{\sqrt 2 }}$.

Substituting the value of x in (iii), y² = $\frac{1}{2}$→ y = ± $\frac{1}{{\sqrt 2 }}$.

Since, 2xy = - 5 < 0.

So, x = $\frac{5}{{\sqrt 2 }}$, y = $ - \frac{1}{{\sqrt 2 }}$→$\frac{5}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}$i and x = - $\frac{5}{{\sqrt 2 }}$, y =  $\frac{1}{{\sqrt 2 }}$à$ - \frac{5}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}$i

So, the required roots are i.e. ± $\left( {\frac{5}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}{\rm{i}}} \right)$ = ±$\frac{1}{{\sqrt 2 }}$(5 – i).

 

i) $\frac{{2 - 36{\rm{i}}}}{{2 + 3{\rm{i}}}}$

Solution:

Let x + iy be the square root of  $\frac{{2 - 36{\rm{i}}}}{{2 + 3{\rm{i}}}}$

so that (x + iy)² = $\frac{{2 - 36{\rm{i}}}}{{2 + 3{\rm{i}}}} = \frac{{2 - 36{\rm{i}}}}{{2 + 3{\rm{i}}}}{\rm{*}}\frac{{2 - 3{\rm{i}}}}{{2 - 3{\rm{i}}}}$ = $\frac{{4 - 6{\rm{i}} - 72{\rm{i}} + 108{{\rm{i}}^2}}}{{4 - 9{{\rm{i}}^2}}}$ = $\frac{{4 - 78{\rm{i}} - 108}}{{4 + 9}}$ = $ - \frac{{104}}{{13}} - \frac{{78}}{{13}}{\rm{i}}$ = - 8 – 6i.

Or, x² – y² + 2xy.i = - 8 – 6i.

Equation real and imaginary parts,

Or, x² – y²  = -8…(i)

Or, 2xy = -6 …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = 64 + 36 = 100.

Or, x² + y² = 10 …(iii)

Adding (i) and (iii), x² = 1 → x = ± 1.

Substituting the value of x in (iii), y² = 9 à y = ± 3.

Since, 2xy = - 6 < 0.

So, x = 1, y = - 3. → 1 – 3i  and x = - 1, y = 3 à - 1 + 3i.

So, the required roots are ± (1 – 3i).

 

j)$\frac{{\left( {5,12} \right)}}{{3, - 4}}$ 

Solution:

Let x + iy be the square root of  $\frac{{\left( {5,12} \right)}}{{3, - 4}}$ = $\frac{{5 + 12{\rm{i}}}}{{3 - 4{\rm{i}}}}$.

so that (x + iy)² = $\frac{{5 + 12{\rm{i}}}}{{3 - 4{\rm{i}}}}$ = $\frac{{5 + 12{\rm{i}}}}{{3 - 4{\rm{i}}}}{\rm{*}}\frac{{3 + 4{\rm{i}}}}{{3 + 4{\rm{i}}}}$ = $\frac{{15 + 20{\rm{i}} + 36{\rm{i}} + 48{{\rm{i}}^2}}}{{9 - 16{{\rm{i}}^2}}}$ = $\frac{{15 + 56{\rm{i}} - 48}}{{9 + 16}}$ = $\frac{{ - 33 + 56{\rm{i}}}}{{25}}$ = $ - \frac{{33}}{{25}}{\rm{\: }}$+ $\frac{{56}}{{25}}$i

Or, x² – y² + 2xy.i = $ - \frac{{33}}{{25}} + \frac{{56}}{{25}}{\rm{i}}$

Equation real and imaginary parts,

Or, x² – y²  = $ - \frac{{33}}{{25}}$ …(i)

Or, 2xy = $\frac{{56}}{{25}}$ …(ii)

Again,, (x² + y²)² = (x² – y²)² + (2xy)² = ${\left( { - \frac{{33}}{{25}}} \right)^2} + {\left( {\frac{{56}}{{25}}} \right)^2}$ = $\frac{{1089 + 3136}}{{625}}$ = $\frac{{4225}}{{625}}$$\frac{{169}}{{25}}$

Or, x² + y² = $\frac{{13}}{5}$ …(iii)

Adding (i) and (iii), 2x² = $\frac{{32}}{{25}}$→ x² = $\frac{{16}}{{25}}$→ x = ± $\frac{4}{5}$.

Substituting the value of x in (iii), y² = $\frac{{49}}{{25}}$→y = ± $\frac{7}{5}$.

Since, 2xy = $\frac{{56}}{{25}}$> 0.

So, x = $\frac{4}{5}$, y =${\rm{\: }}\frac{7}{5}$. →$\frac{4}{5} + \frac{7}{5}$i = $\frac{1}{5}$(4 + 7i) and x =$ - \frac{4}{5}$, y = $ - \frac{7}{5}$à$ - \frac{4}{5} - \frac{7}{5}$i = $ - \frac{1}{5}$(4 + 7i).

So, the required roots are ± $\frac{1}{5}$(4 + 7i).