Question

How do you divide `( 2i+5) / ( -7 i + 7 )` in trigonometric form?

Answer 1

`0.54(cos(1.17)+isin(1.17))`

Explanation:

Let's split them up into two separate complex numbers to start with, one being the numerator, `2i+5`, and one the denominator, `-7i+7`.

We want to get them from linear (`x+iy`) form to trigonometric (`r(costheta + isintheta)` where `theta` is the argument and `r` is the modulus.

For `2i+5` we get

`r = sqrt(2^2 + 5^2) = sqrt29`

`tantheta = 2/5 -> theta = arctan(2/5) = 0.38 " rad"`

and for `-7i+7` we get

`r = sqrt((-7)^2+7^2) = 7sqrt2`

Working out the argument for the second one is more difficult, because it has to be between `-pi` and `pi`. We know that `-7i+7` must be in the fourth quadrant, so it will have a negative value from `-pi/2 < theta < 0`.

That means we can figure it out simply by

`-tan(theta) = 7/7 = 1 -> theta = arctan(-1) = -0.79 " rad"`

So now we've got the complex number overall of

`(2i+5)/(-7i+7) = (sqrt29(cos(0.38)+isin(0.38)))/(7sqrt2(cos(-0.79)+isin(-0.79)))`

We know that when we have trigonometric forms, we divide the moduli and subtract the arguments, so we end up with

`z = (sqrt29/(7sqrt2))(cos(0.38+0.79)+isin(0.38+0.79))`

`= 0.54(cos(1.17)+isin(1.17))`