Euler form of complex number MCQ - Practice Questions & Answers

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Euler form of complex number, Properties of argument of a complex number is considered one of the most asked concept.
21 Questions around this concept.

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

$3\pi/4$

$\pi/2$

$\pi/4$

$5\pi/4$

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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^iΘ. So, z = r (cos ? + i sin ?) is written as r.e^iΘ in Euler's Form

We know the expansion of e^x is

$\\\text{The expansion of } e^x \,\,\text{is}\\\\\mathrm{e^x = 1 + \frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.....} \\\\\mathrm{Replacing \;x\; by \; ix } \\\\\mathrm{e^{ix} = 1+\frac{ix}{1!}+\frac{(ix)^2}{2!} + \frac{(ix)^3}{3!}+ \frac{(ix)^4}{4!}+.... } \\\\\mathrm{e^{ix} = 1 + \frac{ix}{1!}-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+...+} \\\\\mathrm{rearranging\; the \; terms, we\; have} \\\\\mathrm{e^{ix} = (1-\frac{x^2}{2!}+\frac{x^4}{4!}) + i(x-\frac{x^3}{3!}+\frac{x^5}{5!})} \\\\\mathrm{We \; notice\; that\; first \; bracket\; is \; the \; expansion \; of \; \sin x \; } \\\mathrm{and \; 2nd \; bracket\; is \; the \; expansion\; of \;\cos x, so\; we \; have} \\\mathrm{e^{ix}=\sin x +i\cos x}$

So, e^i? = cos? + isin? and

e^-i? = cos? - isin?

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

$\\\mathrm{z=x+iy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\(Cartesian\;form)}\\\mathrm{z=|z(\cos\theta+i\sin\theta)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(Polar\;form)}\\\mathrm{z=|z|e^{i\theta}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\(Eular's\;form)}$

Application of Euler form:

1. Multiplication of two complex numbers:

$\\\mathrm{Let \;z=\left | z \right |e^{i\theta_1}} \\\mathrm{And \;w=\left | w \right |e^{i\theta_2}} \\\mathrm{Multiplying\; these \;two \;number } \\\mathrm{z\cdot w= \left | z \right |e^{i\theta_1}\cdot \left | w \right |e^{i\theta_2}} \\\mathrm{= \left | z \right |\cdot \left | w \right |e^{i(\theta_1+\theta_2)}}$

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

$\\\mathrm{z=|z|e^{i\theta_1}\;\;and\;\;w=|w|e^{i\theta_2}\;be\;two\;complex\;number}\\\mathrm{\therefore\;\frac{z}{w}=\frac{|z|}{|w|}e^{i(\theta_1-\theta_2)}}$

3. The logarithm of Complex Number

$\\\mathrm{z=|z|e^{i\theta}}\\\mathrm{\log_e(z)=\log_e(|z|e^{i\theta})}\\\mathrm{\log_e(z)=\log_e(|z|)+\log_e(e^{i\theta})}\\\mathrm{\log_e(z)=\log_e(|z|)+{i\arg(z)}}$

Properties of argument of a complex number

$\\\mathrm{ i) \,\,arg(z_1z_2)=arg(z_1) + arg(z_2) + 2k\pi, \;\; k\; is \;an\;integer}$

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 complex in similar way

$\\\mathrm{ ii) arg\left ( \frac{z_1}{z_2} \right ) = arg(z_1) - arg(z_2) +2k\pi}$ where k is an integer

$\\\mathrm{iii)\,\, arg( \overline{z}) = - arg(z) }$

$\\\mathrm{iv) arg\left ( z^n \right ) = n\cdot arg(z) +2k\pi}$ k belongs to an integer

$\\\mathrm{(v)\;|z_1+z_2|=|z_1|+|z_2|\Rightarrow arg(z_1)=arg(z_2)}\\\mathrm{(vi)\;|z_1+z_2|=||z_1|-|z_2||\Rightarrow arg(z_1)-arg(z_2)=\pi}$

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