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

2 Answers
A. S. Adikesavan
Jan 12, 2017

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

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.

George C.
Mar 12, 2017

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

Explanation:

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))

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