Answer to Question #214405 in Algebra for prince

Question

(3.1)

(a) Find a and b such that −3ai − (−1 − i)b = 3a − 2bi.

(b) Let z₁ = 12 + 5i and z2 = (3 − 2i)(2 + λi). Find λ without resorting to division such that z₂ = z₁. 

(3.2) Let z = −2 + 3i and z^' = 5 − 4i. Determine the complex numbers

(a) z² − zz'

(b) 1/2(z + z)²

(c) 1/2[z − z] + [(−1 − z^')]².

Answer 1

(3.1)

$(a)\\ -3ai-(-1-i)b=3a-2bi\\ -3ai+b+bi=3a-2bi\\ (-3a+b)i+b=3a-2bi\\ \therefore\\ -3a+b=-2b\\ b=3a\\ -3a+3a=-2b\\ -2b=0\\ b=0\\ a=0\\ (b)\\ z1 = 12 + 5i \\ z2 = (3 − 2i)(2 + λi)=6+3\lambda{i}-4i+2\lambda\\ z_2=z_1\\ 6+3\lambda{i}-4i+2\lambda=12+5i\\ 6+2\lambda=12\\ 3\lambda{}-4=5\\$

For each last two equation

$\lambda=3$

(3.2)

(a)

$z = −2 + 3i\\ z' = 5 − 4i\\ z^2=(-2)^2+2(-2)(3i)+(3i)^2=-5-12i\\ zz'=(-2+3i)(5-4i)=-10+8i+15i+12\\ zz'=2+23i\\ z^2-zz'=-7-35i\\$

(b)

$\frac{1}{2(z+z')^2}=\frac{1}{2(-2+3i+5-4i)^2}=\frac{1}{2(3-i)^2}=\frac{1}{2(8-6i)}=\\\frac{1}{16-12i}\\$

Rationalizing

$\frac{16+12i}{(16+12i)(16-12i)}\\ \frac{16+12i}{(16+12i)(16-12i)}\\ \frac{16+12i}{(16^2-(12i)^2)}\\ \frac{16+12i}{(400)}\\ \frac{16+12i}{400}\\ \frac{16}{400}+\frac{12i}{400}\\ \frac{1}{25}+\frac{3i}{100}$

(c)

$\frac{1}{2[z-z']}+[(-1-z')]^2\\ \frac{1}{2[-2+3i-(5-4i)]}+[(-1-(5-4i)]^2\\ \frac{1}{2[-7+7i]}+[-6+4i]^2\\ \frac{1}{-14+14i}+36-48i-16\\ \frac{-14-14i}{(-14+14i)(-14-14i)}+20-48i\\ \frac{-14-14i}{(196+196)}+20-48i\\ \frac{-14-14i}{392}+20-48i\\ \frac{-14}{392}-\frac{14i}{392}+20-48i\\ \frac{-1}{28}-\frac{i}{28}+20-48i\\ \frac{559}{28}-\frac{1345i}{28}$