find all the values of of λ∈c such that λ(1+2i, 5+4i) =(3 +2i, 6 - i)
Given, "\u03bb(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"
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