Question

Consider the expression $a+ib={{\left( 1-i\sqrt{3} \right)}^{100}}$ , where $i=\sqrt{-1}$ then find the value of a and b.

Answer 1

Hint: Find the condition to convert the imaginary into a solvable form like in terms of $\omega$ .By using Euler’s formula of sin and cos of an angle
$Cisx=\cos x+i\sin x$.

Complete step-by-step solution-
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
Given expression:
$a+ib={{\left( 1-i\sqrt{3} \right)}^{100}}$
Take right hand side expression
By Euler’s formula of sin and cos of an angle
$Cisx=\cos x+i\sin x$
Try to convert the term into this form
$={{\left( 1-i\sqrt{3} \right)}^{100}}$
By multiplying and dividing by 2 inside the bracket the term changes into:
$={{\left( \dfrac{1-i\sqrt{3}}{2}\times 2 \right)}^{100}}$
Taking 2 outside the bracket the term changes into:
$={{2}^{100}}{{\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)}^{100}}$
Take the term inside the bracket, we get:
$\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$
Converting it into cos and sin, we get
$\cos \dfrac{5\pi }{3}+i\sin \dfrac{5\pi }{3}=Cis\dfrac{5\pi }{3}$
Converting it into ${{e}^{ix}}$ terms, we get:
${{e}^{i\dfrac{5\pi }{3}}}$
Substituting this back into the equation, we get:
$={{2}^{100}}{{\left( {{e}^{i\dfrac{5\pi }{3}}} \right)}^{100}}={{2}^{100}}{{e}^{i\dfrac{500\pi }{3}}}={{2}^{100}}\left( {{e}^{i\dfrac{498\pi }{3}}}.{{e}^{i\dfrac{2\pi }{3}}} \right)$
By simplifying, we get
$\begin{aligned} & ={{2}^{100}}\left( {{e}^{i\left( 166\pi \right)}}.{{e}^{i\dfrac{2\pi }{3}}} \right) \\\ & ={{2}^{100}}\left( \left( \cos \left( 166\pi \right)+i\sin \left( 166\pi \right) \right)\left( \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right) \right) \\\ \end{aligned}$
We know

& \cos \left( 166\pi \right)=1 \\\ & \sin \left( 166\pi \right)=0 \\\ & \cos \left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2} \\\ & \sin \left( \dfrac{2\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ By substituting these, we get $\begin{aligned} & ={{2}^{100}}\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right) \\\ & =-{{2}^{99}}+{{2}^{99}}\sqrt{3}i \\\ \end{aligned}$ Given, $a+ib={{\left( 1-i\sqrt{3} \right)}^{100}}$ From above, we get: $a+ib=-{{2}^{99}}+{{2}^{99}}\sqrt{3}i$ By comparing, we get: $a=-{{2}^{99}},b={{2}^{99}}\sqrt{3}$. Note: Be careful while converting imaginary numbers into the ${{e}^{ix}}$ form as the whole question depends on it. After getting it into Euler’s form convert into the form of cis (x). By this we can get the value in the terms of cos and sin of angle. We do this conversion as sin and cos can be found easily than exponential form.