Question

(2+11i)+(4-6i)
(2+11i)-(4-6i)
(4-6i)-(2+11i)
(2+11i).(4-6i)
(2+11i)÷(4-6i)​

Answer 1

(2+11i)+(4-6i)=(2+4)+(11-6)i
=(6)+(5)i
=6+ 5i Parte real: 6 Parte imaginaria: 5

(2+11i)-(4-6i)=(2-4)+(11+6)i
=(-2)+(17)i
=-2+17i Parte real: -2 Parte imaginará: 17

(4-6i)-(2+11i)=(4-2)+(-6-11)i
=(2)+(-17)i
=2-17i Parte real: 2 Parte imaginaria: -17

(2+11i)(4-6i)=8-12i+44i-66i^2
=8+32i-66(-1)
=8+32i+66
=74+32i Parte real: 74 Parte imaginaria: 32 (i^2= -1)

2+11i/4-6i=(2+11i/4-6i)(4-6i/4-6i)
=8+12i+44i+66i^2/16+24i-24i-36i^2
=8+56i+66(-1)/16-36(-1)
=8+56i-66/52
= -58+56i/52
= -29/26 +28/26i Parte real: -29/26 Parte imaginaria: 28/26i