Question

find all the values of of λ∈c such that λ(1+2i, 5+4i) =(3 +2i, 6 - i)

Answer 1

Given, $λ(1+2i, 5+4i) =(3 +2i, 6 - i)$

$\lambda(1+2i)=3+2i$

$\lambda=\dfrac{3+2i}{1+2i}=\dfrac{(3+2i)(1-2i)}{(1+2i)(1-2i)}=\dfrac{3-6i+2i-4i^2}{1-4i^2}=\dfrac{7-4i}{5}$

Now,

$\lambda(5+4i)=6-i$

$\lambda=\dfrac{6-i}{5+4i}=\dfrac{(6-i)(5-4i)}{(5+4i)(5-4i)}=\dfrac{30-24i+20i+4i^2}{25-16i^2}=\dfrac{26-4i}{41}$

The values of $\lambda \text{ are }\dfrac{7-4i}{5} \text{ and }\dfrac{26-4i}{41},\text{ where }\lambda\in C$