Question

Chapter :- Complex numbers.

Q) Find the argument and Principle argument of (1-i√3)^9.

Need The Answer Urgently, Huge request Please No spamming. ​

Answer 1

We're asked to find the argument and principal argument of the complex number,

$\small\displaystyle\longrightarrow z=\left(1-i\sqrt3\right)^9$

In the complex number $\small\displaystyle1-i\sqrt3$ the real part is 1 and imaginary part is -√3.

Here the complex number $\small\displaystyle z$ is first written in Euler form.

First both sides are divided by 9th power of $\small\displaystyle\sqrt{1^2+(-\sqrt3)^2}=2,$ i.e., by $\small\displaystyle 2^9.$

$\small\displaystyle\longrightarrow\dfrac{z}{2^9}=\dfrac{\left(1-i\sqrt3\right)^9}{2^9}$

$\small\displaystyle\longrightarrow\dfrac{z}{2^9}=\left(\dfrac{1}{2}-i\cdot\dfrac{\sqrt3}{2}\right)^9$

Since $\small\displaystyle\cos\left(-\dfrac{\pi}{3}\right)=\dfrac{1}{2}$ and $\small\displaystyle\sin\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt3}{2},$

$\small\displaystyle\longrightarrow\dfrac{z}{2^9}=\left(\cos\left(-\dfrac{\pi}{3}\right)+i\sin\left(-\dfrac{\pi}{3}\right)\right)^9$

Here the complex number $\small\displaystyle1-i\sqrt3$ is in polar form. Then its Euler form will be,

$\small\text{ \displaystyle\longrightarrow\dfrac{z}{2^9}=\left(e^{-i\frac{\pi}{3}}\right)^9 }$

$\small\text{ \displaystyle\longrightarrow z= 2^9e^{-i\frac{9\pi}{3}} }$

$\small\displaystyle\longrightarrow z= 2^9e^{-i\,3\pi}$

Now $\small\displaystyle z$ is in Euler form. It's argument is shown $\small\displaystyle-3\pi$ so the arguments are in the form $\small\displaystyle 2k\pi-3\pi\quad\!\forall k\in\mathbb{Z}.$

And the principal argument is given by the value of $\small\displaystyle k$ such that,

$\small\displaystyle\longrightarrow2k\pi-3\pi\in(-\pi,\ \pi]$

$\small\displaystyle\longrightarrow2k\pi\in(2\pi,\ 4\pi]$

$\small\displaystyle\longrightarrow k\in(1,\ 2]$

Since $\small\displaystyle k\in\mathbb{Z},$

$\small\displaystyle\longrightarrow k=2$

Then principal argument is,

$\small\displaystyle\longrightarrow2k\pi-3\pi=2(2)\pi-3\pi=\pi$

Hence the principal argument is $\small\displaystyle\pi$ and so the general argument is $\small\displaystyle 2n\pi+\pi\quad\!\forall n\in\mathbb{Z}.$