<p>Let \[z^n = r\left[\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)\right].\] Then \[z = r^{\frac{1}{n}}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right], k = 0, 1, 2, \ldots, n - 1.\]</p>
<p>Note :</p>
<ol class="i-list"><li><div class="list-item-paragraph">There are \(n\) distinct \(n^{\mathrm{th}}\)_ roots of \(r(\cos\theta + i\sin\theta)\)</div></li><li><div class="list-item-paragraph">Each root has modulus \(r^{\frac{1}{n}}\).</div></li><li><div class="list-item-paragraph">One of the roots has argument \(\frac{\theta}{n}\).</div></li><li><div class="list-item-paragraph">Successive arguments differ by \(\frac{2\pi}{n}\).</div></li><li><div class="list-item-paragraph">The sum of the roots of a complex number is \(0\)</div></li><li><div class="list-item-paragraph">On an Argand diagram, the points representing the \(n^{\mathrm{th}}\)_ roots of \(r(\cos\theta + i\sin\theta)\) lie on a circle of radius \(r^{\frac{1}{n}}\)_ and are separated by angles \(\frac{2\pi}{n}\).</div></li><li><div class="list-item-paragraph">Unless \(\theta\) is a multiple of \(\pi\), the roots do not occur in conjugate pairs.</div></li></ol>

Let \[z^n = r\left[\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)\right].\] Then \[z = r^{\frac{1}{n}}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right], k = 0, 1, 2, \ldots, n - 1.\]

There are \(n\) distinct \(n^{\mathrm{th}}\)_ roots of \(r(\cos\theta + i\sin\theta)\)

Each root has modulus \(r^{\frac{1}{n}}\).

One of the roots has argument \(\frac{\theta}{n}\).

Successive arguments differ by \(\frac{2\pi}{n}\).

The sum of the roots of a complex number is \(0\)

On an Argand diagram, the points representing the \(n^{\mathrm{th}}\)_ roots of \(r(\cos\theta + i\sin\theta)\) lie on a circle of radius \(r^{\frac{1}{n}}\)_ and are separated by angles \(\frac{2\pi}{n}\).

Unless \(\theta\) is a multiple of \(\pi\), the roots do not occur in conjugate pairs.

Roots of a complex number

Roots of a complex number