Complex numbers and square roots?

Question

How would you solve sqrt(4i + 3) in the form a + bi? I haven't been exposed enough to complex numbers to know how to solve it.

Answer 1

Let a+ib be the square root
(a+ib)^2=a^2+2ab*i-b^2 = a^2-b^2+2a*b*i = 3 + 4i

Comparing Real and imaginary parts we get:
a^2-b^2=3....(1)
and
2ab=4
ab=2
b=2/a
Substitute in equation (1)
a^2-4/a^2=3
a^4-3a^2-4=0
(a^2-4)(a^2+1)=0
a^2+1=0 gives an extraneous root
so,
a^2-4=0
a=+/-2
b=+/-1
NOTE that +/- for both a and b are in their mentione order repectively.
a + bi= +/-(2+i)

Thus sqrt(3+4i) = +/-(2+ i)

Answer 2

The square root of an imaginary number is another imaginary number. First note that (a+bi)^2 = a^2 +2abi -b^2 =
a^2-b^2 + (2ab)i

So for example, (2+i)^2 = 4 +4i -1) =3+4i
Therefore the sqrt of 3+4i is 2+i and 2-1 since complex numbers always come in conjugate pairs.

So let's see how to show that the sqrt of 3+4i is indeed 2 + i and 2 - i

Remember (a+bi)^2= a^2 -b^2 + (2ab)i
So we must show that a^2 - b^2 = 3 and 2ab = 4
Put b= 2/.a into first equation getting a^2-4/a^4=3
a^4 -3a^2 -4 =0This is a quadratic in a^2 which can be solve by factoring giving a^2 = 4 or -1. Reject the -1 as we want a real solution so x^2 = 4 and a = 2 or -2
b=2/a so b= 1 or -1
Hence the sqrt of 3+=4i is 2+i or 2-i

Hope this helped.

Answer 3

Another way is to use polar coordinates. 3 + 4i has absolute value of 5 (3-4-5 right triangle), so the sin of the angle is 4/5.

For a number in polar coordinates (r, a) the square root is (sqrt(r), a/2) so you just need to use the half angle formula to get the sin and cos of arcsin(4/5)/2 = b, so then you have sqrt(5)cos(b) + i*sqrt(5)sin(b) = sqrt(3 + 4i)

Answer 4

let sqrt(3+4i) be a+ib where a and b are real

(a+ib)^2 = 3 + 4i

or a^-b^2+2abi = 3 + 4i

equate the real and imaginary on both sides

a^2-b^2 = 3

2ab = 4 or ab = 2..1

try a = 2 and b =1 and it matches the 1st equation also

so square root = 2 + i

if it did not match then we can solve it through some means

Answer 5

z = a+bi = sqrt(4i+3), square everything
a² + 2abi + b²i² = 4i+3

real parts have to be equal ---> a²- b² = 3
imaginary parts equal --> 2abi = 4i ---> b=2/a

Solving for a and b, we get a=2; b=1 or a=-2; b=-1.

So the roots are ±(2+i).

Checking: (±(2+i))² = 4 + 4i + i² = 4i+3.

Answer 6

Hey Puggy. This time I got here on time to answer your question before you close it.

I only use the polar form to square or multiply or divide complex numbers. It only requires a bit practice in the passage polar-binomial form and viceversa.

In this case its pretty easy, since the module is 5 and the argument is arct 4/3 (aprox 54o)

Of course you can do things like other persons told you.

Take care

Anabel

Answer 7

let a+ib=sq.rt.(4i+3)
squaring both sides
a^2-b^2+2aib=4i+3
a^2-b^2=3
and 2ab=4
ab=2
a=b/2 or b=a/2
sub b=a/2
a^2-(a^2/4)=3
3a^2/4=3
a^2=4
a=+/-2
b=+/-1
so sq.rt.(4i+3)=+/-(2+i)
suppose we put a=b/2
b^2/4-b^2=3
-b^2=4
b=+/-2i
so sq.rt a+ib=+/-(i+2)

Answer 8

i havent been exposed enough etheir