Lösung 3.1:2d

Aus Online Mathematik Brückenkurs 2

Version vom 10:31, 11. Mär. 2009

Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator,

5−11+i3i+i2−3i=1+i5(1+i)−11+i=1+i5+5i−1=1+i4+5i=2−3i3i(2−3i)+i2−3i=2−3i6i−9i2+i=2−3i9+7i.

Hence,

We multiply out the numerator and denominator

(9+7i)(1+i)(4+5i)(2−3i)=91+9i+7i1+7ii42−43i+5i2−5i3i=9+9i+7i−78−12i+10i+15=2+16i23−2i.

This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator,

2+16i23−2i=(2+16i)(2−16i)(23−2i)(2−16i)=22−(16i)2232−2316i−2i2+2i16i=4+25646−368i−4i−32=26014−372i.

If we divide up the numbers into factors,

14372260=27=2186=2293=22331=1026=25132

we can simplify the answers

14260−260372i=2722513−2251322331i=72513−513331i=7130−6593i.