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In complex numbers √-a * √-b = ?

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Answered by Anonymous

to find complex numbers a and b for which the equality

to find complex numbers a and b for which the equality √ab=✓a✓b(✓z)

being principal square root)holds. I tried as below

being principal square root)holds. I tried as below√

being principal square root)holds. I tried as below√ab

being principal square root)holds. I tried as below√ab=

being principal square root)holds. I tried as below√ab=√

being principal square root)holds. I tried as below√ab=√a

being principal square root)holds. I tried as below√ab=√a√

being principal square root)holds. I tried as below√ab=√a√b

being principal square root)holds. I tried as below√ab=√a√b⟺

being principal square root)holds. I tried as below√ab=√a√b⟺√

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)2

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)2 ))

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)2 ))⟺

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)2 ))⟺Arg(ab)=Arg(a)+Arg(b)

being principal square root)holds. I tried as below√ab=√a√b⟺√|ab|(cos(Arg(ab)2 )+isin(Arg(ab)2 )=√|a|√|b|(cos(Arg(a)+Arg(b)2 )+isin(Arg(a)+Arg(b)2 ))⟺Arg(ab)=Arg(a)+Arg(b)Am i right up to this? And can i conclude from above that

is true for all positive reals as in this case Arg(ab)=Arg(a)+Arg(b)=0, for one number being positive real and other negative reals as in this case Arg(ab)=Arg(a)+Arg(b)=π? And for both number being negative reals then Arg(ab)=0 and Arg(a)+Arg(b)=π+π=2π. 2π=0

✓ab=✓a✓b holds.

hope it will help you mark as brainlist answer

Answered by hetalpatel4121982

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In complex numbers √-a * √-b = ?