How do you convert #sqrt(3+4i) # to polar form?

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Answer 1 · 2017-01-12T05:34:35

#+-sqrt5 cis(26.565^o)#, nearly. See explanation.

Only for real #a^2#, sqrta^2)=|a| #, by convention ( or definition ).

So, the other square root #-|a| # is out of reach.

Here, for sqrt(z), where z is complex, we cannot conveniently take

one and keep off the other.

So, #sqrt(3+4i)=(3+i4)^(1/2)# that has two values.

They are #sqrt5(cos a+isina)^(1/2)#,

where #cos a =3/5 and sin a = 4/5#

#=sqrt5(cos(a+2kpi)+isin(a+2kpi))^(1/2), k =0, 1#

#sqrt5(cos(a/2+kpi)+isin(a/2+kpi)), k =0, 1#,

using De Moivre's theorem

#=sqrt5(cos (a/2)+isin(a/2)) and sqrt5(cos (a/2+pi)+isin(a/2+pi))#

#=+-sqrt5(cos (a/2)+isin(a/2))#

#=sqrt5cis(26.565^o)#, nearly.

using #cos (pi+theta)=-cos theta and sin (pi+theta)=-sin theta#.

Answer 2 · 2017-03-12T18:44:00

#sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))#

Note that:

#(2+i)^2 = 4+4i+i^2 = 3+4i#

Since #3+4i# is in Q1, #2+i# is its principal square root.

Further note that:

#abs(2+i) = sqrt(2^2+1^2) = sqrt(5)#

So we have:

#sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))#