Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 20/Solution 1

The absolute value (or modulus) of a complex number $a+bi$ is

$r={\sqrt {a^{2}+b^{2}}}={\sqrt {(-1)^{2}+1^{2}}}={\sqrt {2}}$

and the argument, $\theta$, is the angle between the right side of the $x$-axis and the position of the complex number on the plane. In this case, the argument is $3\pi /4$. To see that, you can either just visually see that the point (-1,1) is clearly on the diagonal of the second quadrant of the plane, or using trigonometry have

$\cos(\theta )={\frac {a}{r}}={\frac {-1}{\sqrt {2}}}\quad {\text{and}}\quad \sin(\theta )={\frac {b}{r}}={\frac {1}{\sqrt {2}}}$

and from there notice that the corresponding angle is $3\pi /4$ (or in real life compute the angle using a calculator since you only know a very few values of sine and cosine).

So this allows us to write our complex number in polar form as

$-1+i=re^{\theta i}={\sqrt {2}}e^{3\pi i/4}$