Question

How do you write the complex number in standard form `[3/2(cos(pi/2)+isin(pi/2))][6(cos(pi/4)+isin(pi/4))]`?

Answer 1

`-9sqrt(2)/2 + 9sqrt(2)/2i`

Explanation:

Given:

`[3/2(cos(pi/2) + isin(pi/2))][6(cos(pi/4) + isin(pi/4))] =`

Multiply the magnitudes and add the angles:

`9(cos((3pi)/4) + isin((3pi)/4)) =`

To obtain standard form, you evaluate the trig functions and use the distributive property with the magnitude:

`-9sqrt(2)/2 + 9sqrt(2)/2i`