(3.1)
(a)−3ai−(−1−i)b=3a−2bi−3ai+b+bi=3a−2bi(−3a+b)i+b=3a−2bi∴−3a+b=−2bb=3a−3a+3a=−2b−2b=0b=0a=0(b)z1=12+5iz2=(3−2i)(2+λi)=6+3λi−4i+2λz2=z16+3λi−4i+2λ=12+5i6+2λ=123λ−4=5
For each last two equation
λ=3
(3.2)
(a)
z=−2+3iz′=5−4iz2=(−2)2+2(−2)(3i)+(3i)2=−5−12izz′=(−2+3i)(5−4i)=−10+8i+15i+12zz′=2+23iz2−zz′=−7−35i
(b)
2(z+z′)21=2(−2+3i+5−4i)21=2(3−i)21=2(8−6i)1=16−12i1
Rationalizing
(16+12i)(16−12i)16+12i(16+12i)(16−12i)16+12i(162−(12i)2)16+12i(400)16+12i40016+12i40016+40012i251+1003i
(c)
2[z−z′]1+[(−1−z′)]22[−2+3i−(5−4i)]1+[(−1−(5−4i)]22[−7+7i]1+[−6+4i]2−14+14i1+36−48i−16(−14+14i)(−14−14i)−14−14i+20−48i(196+196)−14−14i+20−48i392−14−14i+20−48i392−14−39214i+20−48i28−1−28i+20−48i28559−281345i
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