Question

How do you divide `( i-5) / (2i+12)` in trigonometric form?

Answer 1

`z_1/z_2=sqrt26/(2sqrt37)*cis(159.23˚)`

`=sqrt26/(2sqrt37)(cos(159.23˚)+isin(159.23˚))`

Explanation:

Let

`z_1=-5+i`

`z_2=12+2i`

These are in cartesian form. We need to convert them to polar/trigonometric form. To do this we need the moduli and arguments of each complex number. The modulus is the length of the vector representation and is found using the Pythagorean theorem. The argument is the angle made between the positive x-axis and the vector. This is found using SOH CAH TOA. Hopefully you are familiar with Argand diagrams and the vector representation of complex numbers.

Moduli:

`|z_1|=sqrt((-5)^2+1^2)=sqrt26`

`|z_2|=sqrt(12^2+2^2)=sqrt148`

Arguments:

`cos(theta_1)=(Re(z_1))/|z_1|=-5/sqrt26rArrtheta_1=cos^-1(-5/sqrt26)=168.69˚`

`cos(theta_2)=(Re(z_2))/|z_2|=12/sqrt148rArrtheta_2=cos^-1(12/sqrt148)=9.46˚`

Polar form:

`z_1=|z_1|(cos(theta_1)+isin(theta_1))=sqrt26*cis(168.69˚)`

`z_2=|z_2|(cos(theta_2)+isin(theta_2))=sqrt148*cis(9.46˚)`

Division of complex numbers in polar form is as follows. Divide the moduli, subtract the arguments, then multiply these two parts together:

`z_1/z_2=|z_1|/|z_2|*(cis(theta_1))/(cis(theta_2))=|z_1|/|z_2|*cis(theta_1-theta_2)`

`rArrz_1/z_2=sqrt26/sqrt148*cis(168.69-9.46)=sqrt26/(2sqrt37)*cis(159.23˚)`