Express i*(1+i) in polar form

In this tutorial, I will solve a very simple problem by expressing the complex number \(i*(1+i)\) in polar form, or in exponential form, or Euler form. This is the different names of the same type of representation.

First of all, we need to convert our expression to a canonical complex number form \(a + ib\).

\[i*(1+i) = i + i^2 = i - 1 = -1 + i\]

Then we can calculate standard parameters

\[r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\]

Then we need to find \(Arg(-1, 1)\). I will find it via trigonometrical form of complex numbers \(z = r\times(\cos(\theta) + i\sin(\theta))\):

\[-1 + i = \sqrt{2}\times(x + iy) \Rightarrow -1 = \sqrt{2}x; 1 = \sqrt{2}y \Rightarrow x = \frac{1}{-\sqrt{2}}; y = \frac{1}{\sqrt{2}}\]

\[-1 + i = \sqrt{2}(\frac{1}{-\sqrt{2}} + i\frac{1}{\sqrt{2}})\]

Comaring with the trigonometrical form of complex number, we can write:

\[\cos(\theta) = \frac{1}{-\sqrt{2}}; \sin(\theta) = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{3\pi}{4} + 2\pi n\]

therefore:

\[-1 + i = \sqrt{2}\times(\cos(\frac{3\pi}{4} + 2\pi n) + i\sin(\frac{3\pi}{4} + 2\pi n)) \equiv \sqrt{2}\times e^{i(\frac{3\pi}{4} + 2\pi n)}\]

with principal value:

\[-1 + i = \sqrt{2}\times e^{i\frac{3\pi}{4}}\]

For more detailed explanation, please go to watch our video:

Published: 2023-05-10 02:10:25
Updated: 2023-05-10 02:12:38