Calculate a 1 4i b 1 i3 i2 3i c i1 4i33 2i2 d pi i

Calculate a. |1 + 4i| = b. |(1 + i)(3 - i)(2 - 3i)| = c. |i(1 + 4i)^3/(-3 - 2i)^2| = d. |(pi + i)^96/(pi - i)^96| =

a) . first we have to make everyhting in the form of a+bi then we can find | a+ib| value

given

| (1+4i) / (-3 -i) |

the conjugate of -3-i is -3 +i

now divide and multiply with -3+i

| (1+4i) (-3+i) / (-3-i) (-3+i) | = | (-3+i -12i+4i^2) / (9 -i^2) |

= | (-3 -11i -4 / (9+1) | since (i^2 =-1)

= | -7 -11i/10) |

= sqrt(7^2 +11^2) /10

= sqrt(49 +121) /10

= sqrt(170)/10

b) . (1+i)\' . (3 -i) (2 -3i)

here (1+i)\' is the congugate of 1+i is 1-i

so (1-i) (3-i)(2-3i) = (1-i).(6-9i-2i+3)

=(1-i)(9-11i)

=(9-11i-9i+11)

= (20-20i)

| 20 -20i| = sqrt(400+400) = sqrt(800)

= 20 sqrt(2)

similerly we can solve remaining two

3). |

|i(1+4i)^3) / (-3-2i)^2 |

to solve this first we have to simplify

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

(1+4i)^3 = 1 + 3 (4i) +3 (4i)^2 + (4i)^3

= 1 +12i +3 (-16) -64i

= 1 + 12i -48 -64i

= -47 -52i

(-3 -2i)^2 = 9 +(2i)^2 + 2.3.(2i)

= 9 -4 +12i

= 5 +12i

|i(1+4i)^3) / (-3-2i)^2 | = | i(-47 -52i) / (5+12i) |

= | (-47i +52) /(5+12i) |

= | (-47i+52)(5-12i) /(5+12i) (5-12i)|

= | (-235i -564+260-624i) / (25+144) |

= | (-859i -264)/169|

| (-859i -264)/169| = sqrt(859^2 +264^2)/169

= sqrt(737661 +69696)/169

= sqrt(807357)/169

| (pi +i)^96/ (pi-i)^96