Solve Equation for x & y: 1+z=3-2i

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Given z = x + iy
find x and y if
1 + z = 3 - 2i
I tried subbing x+iy into z and solving but all i got was x = y.
The answer in the back is x=2 y=-2
 
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Separate 1+(x+iy)=3-2i into real and imaginary parts. They have to be equal separately.
 
so is it just x+iy=2-2i ?
 
Ry122 said:
so is it just x+iy=2-2i ?

yeah and from here x=2, y=-2, because like Dick said the real part and the imaginary part on both sides has to be equal. Like
a+bi=c+di<=>a=c, \ \ and \ \ b=d
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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