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$$ f:\mathbb{C}\rightarrow\mathbb{C} $$ $$ f(z) = z^2+6\cdot z+19 $$ $$and$$ $$ \Re(z) > -3 $$ $$ Find \ Im(f) $$ where $Im(f) = \{y\in\mathbb{C} | \exists x \in\mathbb{C}, f(x)=y \}$ (the image of function)

How can I do this? Thank you very much!

I know how to do this with a quadratic equation when there are real numbers, $x\in\mathbb{R}$ (when a<0 or a>0). But when there are complex numbers how can I do this?

Anyone? Maybe some ideas? Thank you very much!!!

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    Anyone? Maybe some ideas? Thank you very much!!! – 2017-01-28
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    Is that $(z^2+6)(z+19)$.? – 2017-01-28
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    @MyGlasses itis $z^2 + 6z +19$ – 2017-01-28

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Let $z=x+iy$ thus $$f(z)=z^2+6z+19=(x^2-y^2+6x+19)+i(2xy+6y)=u+iv$$ with $u=(x+3)^2-y^2+10$ and $v=2y(x+3)$ we may delete $y$ between them, this becomes $$u=(x+3)^2-\frac{v^2}{2(x+3)^2}+10$$ it's simple that take $(x+3)^2=t$ and the equation of image will be $\color{blue}{u=-\dfrac{1}{2t}v^2+t-10}$ where $t>0$. The shape of $f$ is a parabola with vertex $(-10,0)$ like this:

enter image description here

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    @MMPP Good luck. – 2017-01-28