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What are x and y? y/1+i = x^2 + 4/x+2i

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What are x and y?

y/1+i = x^2 + 4/x+2i

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Posted (Answer #1)

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Comparing real and imaginary parts



Thus we have



Solving above biquadratic equation, we have two real roots and other two roots are complex. Since x can not be complex so ignore them. Real roots are

`x approx.93`


So `yapprox 3.29`  when `x approx .93`  and

`yapprox2.56` when `xapprox-1.50` 

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