how do you solve (7+24i)/(x+iy)^2 = 1? i got x= 4i or x=-4i or x=3 or x=-3... but the answer is x =4, y=3 or x=-4, y=-3 thanks!
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Originally Posted by alexandrabel90 how do you solve (7+24i)/(x+iy)^2 = 1? i got x= 4i or x=-4i or x=3 or x=-3... but the answer is x =4, y=3 or x=-4, y=-3 thanks! $\displaystyle 7 + 24i = x^2 - y^2 + 2xy i$. Therefore: $\displaystyle 7 = x^2 - y^2$ .... (1) $\displaystyle 24 = 2xy$ .... (2) Solve simultaneously for x and y.
Originally Posted by alexandrabel90 how do you solve (7+24i)/(x+iy)^2 = 1? i got x= 4i or x=-4i or x=3 or x=-3... but the answer is x =4, y=3 or x=-4, y=-3 thanks! $\displaystyle \frac{7+24i}{(x+iy)^2}=1 \Longrightarrow x^2-y^2+2xyi=7+24i$ Try to take it from here equating real and imaginary parts and choosing wisely the sign of solutions... Tonio
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