First slide

Algebra of complex numbers

Question

Let $z = a + ib = {re}^{iθ} where a, b, θ \in R$ and $i = \sqrt{- 1}$
then $r = \sqrt{(a^{2} + b^{2})} = | z |$
and $θ = \tan^{- 1} (\frac{b}{a}) = \arg (z)$
now, $| z |^{2} = a^{2} + b^{2} = (a + ib) (a - ib) = z \bar{z}$
$\Rightarrow \frac{1}{z} = \frac{\bar{z}}{| z |^{2}}$
and $|z_{1} z_{2} z_{3} \dots . z_{n}| = |z_{1}| |z_{2}| |z_{3}| \dots |z_{n}|$
if $|f (z)| = 1$ then f(z) is called unimodular. In this case
f(z) can always be expressed as $f (z) = e^{iα}, α \in R$
Also, $e^{iα} + e^{iβ} = e^{i (\frac{α + β}{2})} 2 \cos (\frac{α - β}{2})$ and
$e^{iα} - e^{iβ} = e^{i (\frac{α + β}{2})} 2 isin (\frac{α - β}{2})$ where $α, β \in R$

Moderate

Question

If $|z_{1}| = 1, |z_{2}| = 2, |z_{3}| = 3$ and $|z_{1} + z_{2} + z_{3}| = 1$ then $|9 z_{1} z_{2} + 4 z_{3} z_{1} + z_{2} z_{3}|$ is equal to

A

6
B

36
C

216
D

1296

Solution

$\begin{matrix} ∵ |z_{1}| = 1, |z_{2}| = 2 |z_{3}| = 3 \\ ∴ {|z_{1}|}^{2} = 1, {|z_{2}|}^{2} = 4, {|z_{3}|}^{2} = 9 \\ \Rightarrow z_{1} {\bar{z}}_{1} = 1, z_{2} {\bar{z}}_{2} = 4, z_{3} {\bar{z}}_{3} = 9 \\ ∴ \frac{1}{z_{1}} = {\bar{z}}_{1}, \frac{4}{z_{2}} = {\bar{z}}_{2}, \frac{9}{z_{3}} = {\bar{z}}_{3} \end{matrix}$
now,
$\begin{array}{l} |9 z_{1} z_{2} + 4 z_{3} z_{1} + z_{2} z_{3}| \\ = |z_{1} z_{2} z_{3} (\frac{9}{z_{3}} + \frac{4}{z_{2}} + \frac{1}{z_{1}})| = |z_{1}| |z_{2}| |z_{3}| |\frac{9}{z_{3}} + \frac{4}{z_{2}} + \frac{1}{z_{1}}| \\ = |z_{1}| |z_{2}| |z_{3}| |{\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3}| \\ = |z_{1}| |z_{2}| |z_{3}| |z_{1} + z_{2} + z_{3}| \\ \begin{aligned} = |z_{1}| |z_{2}| |z_{3}| |z_{1} + z_{2} + z_{3}| = 1 \cdot 2 \cdot 3 \cdot 1 = 6 \\ [∵ |z_{1} + z_{2} + z_{3}| = 1] \end{aligned} \end{array}$

Question

The value of $\tan (i \ln (\frac{a - ib}{a + ib}))$ $(where i = \sqrt{- 1})$ is equal to

A

$\frac{2 ab}{b^{2} - a^{2}}$
B

$\frac{2 ab}{a^{2} - b^{2}}$
C

$\frac{a^{2} + b^{2}}{a^{2} - b^{2}}$
D

$\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Solution

$\begin{array}{l} |\frac{a - ib}{a + ib}| = \frac{| a - ib |}{| a + ib |} = \frac{\sqrt{(a^{2} + b^{2})}}{\sqrt{(a^{2} + b^{2})}} = 1 \\ ∴ Let \frac{a - ib}{a + ib} = e^{iα} \Rightarrow \frac{(a - ib)^{2}}{a^{2} + b^{2}} = e^{iα} \\ \Rightarrow (\frac{a^{2} - b^{2}}{a^{2} + b^{2}}) - i (\frac{2 ab}{a^{2} + b^{2}}) = e^{iα} \\ \Rightarrow \ln (\frac{a - ib}{a + ib}) = iα \\ ∴ \tan (iln (\frac{a - ib}{a + ib})) = \tan (i \cdot iα) = \tan (- α) \\ = - \tan α = - \frac{\sin α}{\cos α} \\ = \frac{2 ab}{(a^{2} - b^{2})} \end{array}$

Question

If $| z - 3 i | = 3$ $(where i = \sqrt{- 1})$ and $\arg z \in (0, π / 2)$ then $\cot (\arg z) - \frac{6}{z}$ is equal to

A

0
B

-i
C

i
D

$π$

Solution

$\begin{array}{l} ∵ | z - 3 i | = 3 \\ ∴ z - 3 i = 3 e^{iα} \\ \Rightarrow z = 3 i + 3 e^{iα} = 3 (e^{iπ / 2} + e^{iα}) \\ = 3 e^{i (\frac{π}{4} + \frac{α}{2})} \cdot 2 \cos (\frac{π}{4} - \frac{α}{2}) \\ ∴ \arg (z) = (\frac{π}{4} + \frac{α}{2}) \end{array}$
now,
$\begin{array}{l} \cot (\arg z) - \frac{6}{z} = \cot (\frac{π}{4} + \frac{α}{2}) - \frac{6}{6 e^{(\frac{π}{4} + \frac{α}{2})} \cos (\frac{π}{4} - \frac{α}{2})} \\ = \cot (\frac{π}{4} + \frac{α}{2}) - \frac{e^{- 1} (\frac{π}{4} + \frac{α}{2})}{\cos (\frac{π}{2} - (\frac{π}{4} + \frac{α}{2}))} \\ = \cot (\frac{π}{4} + \frac{α}{2}) - \frac{\{\cos (\frac{π}{4} + \frac{α}{2}) - isin (\frac{π}{4} + \frac{α}{2})\}}{\sin (\frac{π}{4} + \frac{α}{2})\}} \\ = \cot (\frac{π}{4} + \frac{α}{2}) - \cot (\frac{π}{4} + \frac{α}{2}) + i \\ = i \end{array}$

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