Question

Problem 4. Find all complex solutions to the equation 26 – 5iz² = 6

Answer 1

1 76_siz36 > Z6-5123 - 6 - 0 26 - ziz3-ziz3_6=0 Z3 (- 3i) - zi (z3 - 3i) -0. (23-3i)(23-31) = 0 - (23- 2i)=0 (z 3 – 3i)=0 23- ai → z (cos 30+ isin 30), 30 - JI+2KTT for k=o 2-2 (cos II tison II) z = gys (cos Tel + izan tud) = 2 ( 5+) K=Og lg 2 tok=0 S for kal 30 = 1 + 2 + ** en + 25 = 314.4" TITLE Zo ga s cos (2015) + i synä t. )]

for k=2 30 - T +4T Now put z3_31=0 = 73=3i so Z=33 (cos 30 + isin 30) (it where 30 = I + 2KIT for k=0 & 30 = I + 0 = 1 K= 0, 1, 2 KEL, 30-1+34 70- + 2T τ π- ι

<= 5" (cos (6-14 ) + isin ( - E)]. for k= 2 , 30 = It + uh .. X=3%[cos (THL) + i sin 1775] - 3% (-cost tignI) or do so all 6 roots are : as (9 tud), 2% (731), -12%, 3% (+ $1), 35.(492439), aki