Chapter 5.7 - Miscellaneous Examples

In the previous section, we completed a discussion on quadratic equations. We saw some miscellaneous examples also. In this section, we will see a few more miscellaneous examples.

Solved example 5.13
If $x+yi~=~\frac{a+bi}{a-bi}$, prove that $x^2+y^2=1$
Solution:
1. We will change the right side into the general form:
$\begin{array}{ll} \frac{a+bi}{a-bi}&{}={}&\frac{a+bi}{a-bi} × \frac{a+bi}{a+bi}& {} &{} \\ \phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2+2abi-b^2}{a^2+b^2}&{} \\ \phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2-b^2}{a^2+b^2}~+~\frac{2abi}{a^2+b^2}&{} \\ \phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2-b^2}{a^2+b^2}~+~\left(\frac{2ab}{a^2+b^2}\right)i&{} \\ \end{array}$
2. So the given expression becomes: $x+yi~=~\frac{a^2-b^2}{a^2+b^2}~+~\left(\frac{2ab}{a^2+b^2}\right)i$
3. Equating the corresponding real and imaginary parts, we get:
$x~=~\frac{a^2-b^2}{a^2+b^2}$
$y~=~\frac{2ab}{a^2+b^2}$
4. Now we can find $x^2+y^2$:
$\begin{array}{ll} x^2+y^2&{}={}&\left(\frac{a^2-b^2}{a^2+b^2} \right)^2~+~\left(\frac{2ab}{a^2+b^2} \right)^2& {} &{} \\ \phantom{x^2+y^2}&{}={}&\frac{(a^2-b^2)^2~+~(2ab)^2}{(a^2+b^2)^2}&{} \\ \phantom{x^2+y^2}&{}={}&\frac{a^4-2a^2 b^2+b^4~+~4a^2 b^2}{(a^2+b^2)^2}&{} \\ \phantom{x^2+y^2}&{}={}&\frac{a^4+2a^2 b^2+b^4}{(a^2+b^2)^2}&{} \\ \phantom{x^2+y^2}&{}={}&\frac{(a^2+b^2)^2}{(a^2+b^2)^2}&{} \\ \phantom{x^2+y^2}&{}={}&1&{} \\ \end{array}$

Solved example 5.14
Find real 𝜽 such that $\frac{3+2i\sin \theta}{1-2i\sin \theta}$ is purely real
Solution:
1. We will change the given expression into the general form:
$\begin{array}{ll} \frac{3+2i\sin \theta}{1-2i\sin \theta}&{}={}&\frac{3+2i\sin \theta}{1-2i\sin \theta}~ × ~\frac{1+2i\sin \theta}{1+2i\sin \theta}& {} &{} \\ \phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3+6i \sin \theta +2i \sin \theta - 4 \sin^2 \theta}{1+4 \sin^2 \theta}&{} \\ \phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3+8i \sin \theta - 4 \sin^2 \theta}{1+4 \sin^2 \theta}&{} \\ \phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\frac{8i \sin \theta}{1+4 \sin^2 \theta}&{} \\ \phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\left(\frac{8 \sin \theta}{1+4 \sin^2 \theta}\right)i&{} \\ \end{array}$
2. So the given complex number is: $\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\left(\frac{8 \sin \theta}{1+4 \sin^2 \theta}\right)i$
3. If this complex number is to be purely real, the imaginary part must be zero.
• That is., $\frac{8 \sin \theta}{1+4 \sin^2 \theta}$ must be zero.
• Denominator cannot be zero because, division by zero will give a number which does not exist.
• So we can write:
If the given complex number is to be purely real, $8 \sin \theta$ must be zero.
   ♦ That is., $8 \sin \theta~=~0$
   ♦ This is true only when sin 𝜽 = 0
4. sin 𝜽 = 0 is a trigonometrical equation.
• We know that, this equation is true whenever 𝜽 is a multiple of 𝞹
5. So we can write:
The given complex number will be purely real when 𝜽 = n𝞹
   ♦ Where n is member of the set of integers Z.

Solved example 5.15
Convert the complex number $\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ in the polar form.
Solution:
1. We will change the given expression into the general form:
$\begin{array}{ll} \frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}&{}={}&\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}~ × ~\frac{\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}}{\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}}& {} &{} \\ \phantom{\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}}&{}={}&\frac{(i-1)(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3})}{\cos^2 \frac{\pi}{3}+ \sin^2 \frac{\pi}{3}}&{} \\ \phantom{\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}}&{}={}&\frac{(i-1)( \frac{1}{2}- \frac{\sqrt{3}}{2}i)}{1}&{} \\ \phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&(i-1)( \frac{1}{2}- \frac{\sqrt{3}}{2}i)&{} \\ \phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&{\frac{1}{2}}i+\frac{\sqrt{3}}{2}-\frac{1}{2}+{\frac{\sqrt{3}}{2}}i&{} \\ \phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&\frac{\sqrt{3}-1}{2}~+~\left(\frac{\sqrt{3}+1}{2} \right)i&{} \\ \end{array}$
2. Now the complex number is in the form x+yi
We have to convert it into the form r[cos 𝜽+ i sin 𝜽]
3. We know that the modulus r is given by: $r=\sqrt{x^2+y^2}$
So in our present case,
$\begin{array}{ll} r&{}={}&\sqrt{x^2+y^2}& {} &{} \\ \phantom{r}&{}={}&\sqrt{\left(\frac{\sqrt{3}-1}{2} \right)^2+\left(\frac{\sqrt{3}+1}{2} \right)^2}&{} \\ \phantom{r}&{}={}&\sqrt{\frac{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2}{4}}&{} \\ \phantom{r}&{}={}&\sqrt{\frac{3-2\sqrt{3}+1~+~3+2\sqrt{3}+1}{4}}&{} \\ \phantom{r}&{}={}&\sqrt{\frac{8}{4}}&{} \\ \phantom{r}&{}={}&\sqrt{2}&{} \\ \end{array}$
4. Since x+yi and r[cos 𝜽+ i sin 𝜽] represent the same complex number, we can equate the corresponding real and imaginary parts. We get:
    ♦ x = r cos 𝜽
    ♦ y = r sin 𝜽
5. From this we get:
(i) $\frac{\sqrt{3}-1}{2}=\sqrt{2} \cos \theta$
(ii) $\frac{\sqrt{3}+1}{2}=\sqrt{2} \sin \theta$
• sin 𝜽 is +ve. cos 𝜽 is also +ve. So it is in the first quadrant.
6. Taking ratios, (ii) to (i), we get:
$\frac{\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta}=\frac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}}$
$\Rightarrow \tan \theta = \frac{\sqrt{3}+1}{\sqrt{3}-1}$
• This is a trigonometrical equation. We learned to solve them in chapter 3. (Details here)
7. We have, $\tan \frac{5\pi}{12}~=~ \tan 75^o~=~\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Proof:
$\begin{array}{ll} \tan \frac{5 \pi}{12}&{}={}&\tan \left(\frac{5 \pi}{12}~\times ~\frac{2}{2}\right)~=~\tan \frac{10 \pi}{24}& {} &{} \\ \phantom{\tan \frac{5 \pi}{12}}&{}={}&\tan \left(\frac{\pi}{4}+\frac{\pi}{6} \right)&{} \\ \phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{\tan \frac{\pi}{4}~+~\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4}\tan \frac{\pi}{6}}&{} \\ \phantom{\tan \frac{5 \pi}{12}}&{}&\color {green}{\because \tan(p+q)=\frac{\tan p + \tan q}{1-\tan p \tan q}}&{} \\ \phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{1~+~\frac{1}{\sqrt3}}{1-1 \times \frac{1}{\sqrt3}}&{} \\ \phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{\sqrt3+1}{\sqrt3-1}&{} \\ \end{array}$
• From this we get: $\theta = \frac{5\pi}{12}$
• This 𝜽 is in the first quadrant. So there is no need to find the other value of 𝜽.
8. Thus we get the values of r and 𝜽:
   ♦ From (3), we get: $r= \sqrt{2}$
   ♦ From (7),we get: $\theta = \frac{5\pi}{12}$
• So the polar form is: $\sqrt{2}\left[\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right]$
• The red dot in fig.5.9 below represents the complex number:

Fig.5.9

Solved example 5.16
Find the complex number z if $|z+1|=z+2(1+i)$.
Solution:
1. Let z = a+bi. Then the given equation becomes:
$\begin{array}{ll} |a+bi+1|&{}={}&a+bi+2(1+i)& {} &{} \\ \Rightarrow |(a+1)+bi|&{}={}&a+bi+2+2i& {} &{} \\ \Rightarrow \sqrt{(a+1)^2+b^2}&{}={}&a+2+(b+2)i& {} &{} \\ \end{array}$
2. Equating the corresponding real and imaginary parts, we get:
(i) $\sqrt{(a+1)^2+b^2}~=~a+2$
(ii) $0 ~=~b+2$
3. From (ii), we get: b = -2
• Substituting for b in (i), we get:
$\begin{array}{ll} \sqrt{(a+1)^2+b^2}&{}={}&a+2& {} &{} \\ \Rightarrow \sqrt{(a+1)^2+(-2)^2}&{}={}&a+2& {} &{} \\ \Rightarrow \sqrt{a^2+2a+1+4}&{}={}&a+2& {} &{} \\ \Rightarrow \sqrt{a^2+2a+5}&{}={}&a+2& {} &{} \\ \Rightarrow a^2+2a+5&{}={}&(a+2)^2& {} &{} \\ \phantom{\frac{1}{2-3i}}&{}&\color {green}{\text{(Squaring both sides)}} &{} \\ \Rightarrow a^2+2a+5&{}={}&a^2+4a+4& {} &{} \\ \Rightarrow 1&{}={}&2a& {} &{} \\ \Rightarrow a&{}={}&\frac{1}{2}& {} &{} \\ \end{array}$
4. Thus the complex number z = a+bi is $\frac{1}{2}-2i$

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The link below gives a PDF file with more miscellaneous examples

Miscellaneous exercise on chapter 5

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In the next chapter, we will see linear inequalities.

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