How do you add #(2+8i)+(3+4i)# in trigonometric form?

1 Answer
Shwetank Mauria
May 18, 2016

#(2+8i)+(3+4i)=sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]#,

where #alpha=arctan4# and #beta=arctan(4/3)#.

Explanation:

Let us first write #(2+8i)# and #(3+4i)# in trigonometric form.

#a+ib# can be written in trigonometric form #rcosthetaa+irsintheta=r(costhetaa+isintheta)#,
where #r=sqrt(a^2+b^2)# and #tantheta=b/a# or #theta=arctan(b/a)#

Hence #(2+8i)=sqrt(2^2+8^2)[cosalpha+isinalpha]# or

#sqrt68[cosalpha+isinalpha]#, where #alpha=arctan4# and

#(3+4i)=sqrt(3^2+4^2)[cosbeta+isinbeta]# or

#5[cosbeta+issinbeta]#, where #beta=arctan(4/3)#

Hence #(2+8i)+(3+4i)# =

#sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]#,

where #alpha=arctan4# and #beta=arctan(4/3)#.

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