Complex numbers in the form #a+bi# can be represented as:
#z=r(costheta+isintheta)#
Where:
#bbr=sqrt(a^2+b^2)#
#bbtheta=arctan(b/a)#
We now put our complex numbers in this form:
Let:
#z_1=5+4i# and #z_2=6+i#
For #z_1#
#r=sqrt((5)^2+(4)^2)=sqrt(41)#
#theta=arctan(4/5)=38.66#
#z_1=sqrt(41)(cos(38.66)+isin(38.66))#
For #z_2#
#r=sqrt((6)^2+(1)^1)=sqrt(37)#
#theta=arctan(1/6)=9.46#
#z_2=sqrt(37)(cos(9.46)+isin(9.46))#
Addition is the same as for that of complex numbers in the form #a+bi#
#:.#
#a_1+b_1i +a_2+a_2i=(a_1+a_1) +(b_1+b_2)i#
#z_1+z_2#
#sqrt(41)(cos(38.66)+sqrt(37)(cos(9.46)=11.00002714#
#isqrt(41)(sin(38.66)+isqrt(37)sin(9.46)=4.999773551i#
Rounding to 2 .d.p.
#11.00+5.00i=11+5i#
Notice the result is exactly the same as adding in rectangular form.
#(5+4i) +(6+i)=(5+6)+(6+1)i=11+5i#
