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 Math Central Quandaries & Queries
 Question from Douglas: I realize raising 0^a = 0 if a>0 and undefined if a<=0. If have read that 0^bi is undefined for all b. What I don't understand is why 0^(a+bi) = 0 if a and b are not equal to zero. Is this purely by definition or is there a logical reason why this is the case? (I have taken Complex Analysis, so have a fairly good understanding of complex numbers.) Thank you so much for taking the time!

Hi Douglas,

I assume that $a$ and $b$ are real and then

$0^{a + bi} = 0^a \times 0^{bi}.$

If $a \neq 0$ then $0^a = 0$ and hence

$0^{a + bi} = 0^a \times 0^{bi} = 0 \times 0^{bi} = 0$

If $a=0$ then $a+bi = bi$ and hence

$0^{a + bi} = 0^{bi} \mbox { which is undefined.}$

Hence if $a$ and $b$ are real then

$0^{a + bi} = \left\{ \begin{array}{ll} 0 & \mbox{if a \neq 0} \\ \mbox{ undefined} & \mbox{ if a = 0} \end{array} \right.$

Penny

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