division of complex numbers

Question

find the modulus Z=(2-i)(5+12i)/(1+2i)³

Answer 1

z = (2-i)(5+12i)/(1+2i)^3

Now (2-i)(5+12i)

Multiplication in complex numbers (a+bi)(c+di) = (ac-bd)+(bc+ad)i

a = 2, b = -1,c = 5, d = 12.

=(2*5-(-1)*12)+(-1*5+2*12)i

= (10+12)+(-5+24)i

= 22+19i

And now (1+2i)^3 = (1+2i)(1+2i)(1+2i)

(1+2i)(1+2i)

a =1,b = 2,c = 1, d = 2

= (1*1-2*2)+(2*1+1*2)i = -3+4i

(-3+4i)(1+2i)

a = -3,b = 4, c = 1, d = 2

= (-3*1-4*2)+(4*1+-3*2)i

= -11-2i

z = (2-i)(5+12i)/(1+2i)^3

= 22+19i/-11-2i

Division in complex numbers is (a+bi)/(c+di) = [(ac+bd)/c^2+d^2]+[bc-ad/c^2+d^2]

a = 22,b = 19,c =-11,d =-2

= [(22*-11+19*-2)/(-11)^2+(-2)^2]+[(19*-11-22*-2)/(-11)^2+(-2)^2]i

= [-242-38/121+4]+[(-209+44)/121+4]i

z = (-280/125)+(-165/125)i

a = -280/125, b =-165/125

Modulus z = √(a^2+b^2)

= √(-280/125)^2+(-165/125)^2

=√( 2.24^2+1.32^2)

=√(5.01+1.74)

= √6.75

Modulus z = 2.59