How do you add (5+4i)+(6+i) in trigonometric form?

1 Answer
Feb 19, 2018

11+5i

Explanation:

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