Show that the equation Z 4 + 2Z 3 + 3Z 2 + 4Z + 5 = 0 has no root which is either purely real or purely imaginary.
Text Solution
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Sol. Z 4 + 2Z 3 + 3Z 2 + 4Z + 5 = 0
If α is a purely real root of the above equation, then
α 4 + 2 α 3 + 3 α 2 + 4 α + 5 = 0
i.e. ( α 2 + α ) 2 + 2( α + 1) 2 + 3 = 0 which is not possible as all the terms of LHS are positive.
If β is a purely imaginary root of the given equation, then β 4 –2i β 3 –3 β 2 + 4i β + 5 = 0
i.e., β 4 –3 β 2 + 5 = 0 and 2 β 3 + 4 β = 0
i.e. 
+ 
= 0 and 2 β ( β 2 + 2) = 0
Neither of these equations is true. Hence the conclusion.
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