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Euler form of complex number - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Euler form of complex number, Properties of argument of a complex number is considered one of the most asked concept.

  • 20 Questions around this concept.

Solve by difficulty

Let z,\omega  be complex numbers such that  \overline{z}+i\overline{\omega}=0  and  argz\omega =\pi Then arg z equals:

Concepts Covered - 2

Euler form of complex number

The polar form of complex number z = r (cos ? + i sin ?)

In Euler form (cos ? + i sin ?) part of the polar form of complex numbers is represented by e. So, z = r (cos ? + i sin ?) is written as r.e in Euler's Form

We know the expansion of ex is

The expansion of $e^x$ is

$
\mathrm{e}^{\mathrm{x}}=1+\frac{\mathrm{x}}{1!}+\frac{\mathrm{x}^2}{2!}+\frac{\mathrm{x}^3}{3!}+\ldots
$


Replacing x with ix

$
\begin{aligned}
& e^{i x}=1+\frac{i x}{1!}+\frac{(i x)^2}{2!}+\frac{(i x)^3}{3!}+\frac{(i x)^4}{4!}+\ldots \\
& e^{i x}=1+\frac{i x}{1!}-\frac{x^2}{2!}-\frac{i x^3}{3!}+\frac{x^4}{4!}+\ldots+
\end{aligned}
$

rearranging the terms, we have

$
e^{i x}=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)
$


We notice that first bracket is the expansion of $\sin x$ and 2nd bracket is the expansion of $\cos x$, so we have $e^{i x}=\sin x+i \cos x$

 

So, eix = cosx + isinx and 

e-ix = cosx - isinx

 

Euler forms make algebra very simple for complex numbers in cases where multiplication, division or powers of complex numbers are involved. Any complex number can be expressed as

$
\begin{aligned}
& \mathrm{z}=\mathrm{x}+\mathrm{iy} \\
& \mathrm{z}=\mid \mathrm{z}(\cos \theta+\mathrm{i} \sin \theta) \\
& \mathrm{z}=|\mathrm{z}| \mathrm{e}^{i \theta}
\end{aligned}
$

(Cartesian form)
(Polar form)
(Euler's form)

Application of Euler form:
1. Multiplication of two complex numbers:

Let $z=|z| e^{i \theta_1}$
And $\mathrm{w}=|\mathrm{w}| \mathrm{e}^{\mathrm{i} \theta_2}$
Multiplying these two number

$
\begin{aligned}
& \mathrm{z} \cdot \mathrm{w}=|\mathrm{z}| \mathrm{e}^{\mathrm{i} \theta_1} \cdot|\mathrm{w}| \mathrm{e}^{\mathrm{i} \theta_2} \\
& =|\mathrm{z}| \cdot|\mathrm{w}| \mathrm{e}^{\mathrm{i}\left(\theta_1+\theta_2\right)}
\end{aligned}
$
 

2. Division also can be done in the same way,

$\mathrm{z}=|\mathrm{z}| \mathrm{e}^{\mathrm{i} \theta_1}$ and $\mathrm{w}=|\mathrm{w}| \mathrm{e}^{\mathrm{i} \theta_2}$ be two complex number

$
\therefore \frac{\mathrm{z}}{\mathrm{w}}=\frac{|\mathrm{z}|}{|\mathrm{w}|} \mathrm{e}^{\mathrm{i}\left(\theta_1-\theta_2\right)}
$

3. The logarithm of Complex Number

$
\begin{aligned}
& \mathrm{z}=|\mathrm{z}| \mathrm{e}^{\mathrm{i} \theta} \\
& \log _{\mathrm{e}}(\mathrm{z})=\log _{\mathrm{e}}\left(|z| \mathrm{e}^{\mathrm{i} \theta}\right) \\
& \left.\log _{\mathrm{e}}(\mathrm{z})=\log _{\mathrm{e}}|\mathrm{z}|\right)+\log _{\mathrm{e}}\left(\mathrm{e}^{\mathrm{i} \theta}\right) \\
& \log _{\mathrm{e}}(\mathrm{z})=\log _{\mathrm{e}}(|z|)+\mathrm{i} \arg (\mathrm{z})
\end{aligned}
$
 

Properties of argument of a complex number

i) $\arg \left(z_1 z_2\right)=\arg \left(z_1\right)+\arg \left(z_2\right)+2 k \pi, \quad k$ is an integer

Value of $k$ should be chosen in such a way that arg lies between (-?,?]
This formula for argument can be generalised for n number of complexes in a similar way
ii) $\arg \left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)=\arg \left(\mathrm{z}_1\right)-\arg \left(\mathrm{z}_2\right)+2 \mathrm{k} \pi$ where $k$ is an integer
iii) $\arg (\bar{z})=-\arg (z)$
iv) $\arg \left(\mathrm{z}^{\mathrm{n}}\right)=\mathrm{n} \cdot \arg (\mathrm{z})+2 \mathrm{k} \pi$ k belongs to an integer
(v) $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right| \Rightarrow \arg \left(z_1\right)=\arg \left(z_2\right)$
(vi) $\left|z_1+z_2\right|=\left|\left|z_1\right|-\left|z_2\right|\right| \Rightarrow \arg \left(z_1\right)-\arg \left(z_2\right)=\pi$

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Euler form of complex number
Properties of argument of a complex number

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