Question

Find the value of the real number y if the imaginary part and the real part of (1+3i)/(4-iy) are equal.
Please help me solve this problem:) and can u tell me the explanation too pleasee? thankyou soo much!

Answer 1

i get y = -2

Answer 2

how do u get the answerr?

Answer 3

not sure but here is what i did
\[\frac{1+3i}{4-yi}=\frac{1+3i}{4-yi}\times \frac{4+yi}{4+yi}=\frac{-3y+4+(y+12)i}{16+y^2}\]

Answer 4

or
\[\frac{-3y+4}{16+y^2}+\frac{y+12}{16+y^2}i\] and since real and imaginary parts are the same you should have
\[-3y+4=y+12\]

Answer 5

thankyou :)) how about \[\left( 1+3y \right) / \left( 4-iy \right)\]

Answer 6

for that second one:\[y=-\frac{1}{3}\ or\ y=4\]

Answer 7

can you also post the explanations pleasee? i tried to times (4+iy)/(4+iy) to the question but i cant solve ittt :((

Answer 8

if you times it you get \[\frac{4(3y+1)}{y^2+16}+\frac{y(3y+1)}{y^2+16}i\]
since both real and imaginary parts are the same
\[12y+4=3y^2+y\]
solve for y

Answer 9

\[3y^2-11y-4=0\]

Answer 10

thankyou soo muchh:DD