Question

find the square root of z=4+i8​

Answer 1

Firstly, consider the equation x^2 = 9

This has two solutions and we write x = ±√9 = ±3

But when we say “what is the square root of 9”, we really mean

what is √9 and the ONLY answer is 3

The sign √ does not mean that we add a ± sign too.

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So really your question should say “What are the two roots of z^2 = – 4i”

We let z = rcis(θ) and change – 4i to its polar form = 4cis(270 + 360n).

The equation z^2 = – 4i

becomes r^2cis(2θ) = 4cis(270 + 360n)

So r = 2 and 2θ = 270 + 360n so θ = 135 + 180n = 135 or 315 degrees

The two solutions are:

2 cis(135) = –1.414 + 1.414i

and

2cis(315) = +1.414 – 1.414i