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Let $a$ &$ b$ be complex numbers (which may be real) and let, $Z = z^3 + (a + b + 3i) z^2 + (ab + 3 ia + 2 ib - 2) z + 2 abi -2a$

We have to find all purely imaginary numbers $a$ & $b$ when $z = 1 + i$ and $Z$ is a real number .

I tried and factorised it as

Z$=(z+b+i)(z^2+(a+2i)z+2ai)$

But now how to proceed?

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    You want to factor it into $Z = X + Yi$. Then figure out what makes $Y = 0$. To do this, let $z = 1 + i$ in your formula and expand. – 2017-02-01
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    @JoeJohnson126 the answer is given as $(\frac{-2ti}{3t+5} ,ti)$ where t belong to real number except (-5/3) . I could not undrstand how a range can we the answer – 2017-02-01
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    @koolman That’s not the range. They have given the value of $(a,b) $ like that. – 2017-02-01

2 Answers 2

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Putting $z=1+i $ into the expression for $Z $, we get, $$Z =(ab-10-2b-5a) +i (5a + 4b + 3ab) $$ Now as both $a $ and $b $ are imaginary, we can write $a = ki $ and $b= li $ thus simplifying our expression to $$Z = (kl-10-5k-4l) +i (-5k-2l-3kl) $$ A's $Z $ is to be purely real, then the imaginary part should be zero. Thus, $$-5k-2l -3kl =0$$ $$\Rightarrow k = -\frac {2l}{3l +5} $$ From which we get, $$(a,b) = (-\frac {2li}{3l+5}, li) l \neq -\frac {5}{3}$$ Hope it helps.

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I'm writing an answer because the accepted answer is wrong (not just calculation mistakes).

Since $a,b$ are purely imaginary numbers, we can set $a=Ai, b=Bi$ where $A,B\in\mathbb R$.

Then, setting $z=1+i$ gives $$\begin{align}Z&=(1+i+Bi+i)((1+i)^2+(Ai+2i)(1+i)+2\cdot Ai\cdot i)\\\\&=(1+(B+2)i)(-2-3A+(A+4)i)\\\\&=-10-AB-4B-5A+(-2B-3AB-5A)i\end{align}$$ Since $Z$ is real, we have to have $$-2B-3AB-5A=0\iff A(3B+5)=-2B\implies A=\frac{-2B}{3B+5}$$ from which we have $$\color{red}{(a,b)=\left(\frac{-2Bi}{3B+5},Bi\right)}$$ where $B\in\mathbb R$ such that $B\not=-5/3$.