What is the square root of 3 + 4i?

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Let the square root of 3 + 4i be x + iy.

So (x + iy) (x +iy) = 3 +4i

=> x^2 + xyi + xyi + i^2*y^2 = 3 + 4i

=> x^2 – y^2 + 2xyi = 3 + 4i

Equate the real and complex terms

=> x^2 - y^2 = 3 and 2xy = 4

2xy = 4

=> xy = 2

=> x = 2/y

Substitute in

=> x^2 - y^2 = 3

=> 4/y^2 - y^2 = 3

=> 4 – y^4 = 3y^2

=> y^4 + 3y^2 – 4 = 0

=> y^4 + 4y^2 – y^2 – 4 =0

=> y^2(y^2 + 4) – 1(y^2 + 4) =0

=> (y^2 – 1) (y^2 + 4) =0

Therefore y^2 = 1, we ignore y^2 = -4 as it gives complex values of y.

Therefore y = 1 and x = 2/1 = 2

**The required square root of 3 + 4i is 2 + i.**

To find the square root of 3+4i.

Let (3+4i)^(1/2) = p^(1/2)+q^(1/2)*i .

We square both sides:

3+4i = p-q +2(pq)^(1/2)*i

Equating on both side the real parts we get:

p-q = 3....(1)

Similarly equating the imaginary parts, we get:

2(pq)^(1/2) = 4

4pq = 16

pq = 4.

(p+q)^2 = (p-q)^2 +4pq = 3^3+16 = 25.

Therefore p+q = 5....(2)

Therefore (1)+(2) = 2p = 5+3 = 8. So p = 8/2 = 4.

(2)-(1): 2q = 5-3 = 2. So q = 1.

Therefore p^(1/2) = 2(1/2) = 2.

q^(1/2) = (1)^1/2) = 1.

Therefore square root of (4+3i) = 2+i.

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