sqr(3+4i). Express in form a+bi

Answer 1

I presume "sqr" means "square." If not, read on.

sqr(3 + 4i) = (3+4i)(3+4i) = 3*3 + 3*4i + 4i*3 + 4i*4i

= 9 + 12i + 12i + 16*(-1)

= (9 - 16) + 24i

= -7 + 24i

On the other hand, if "sqr" means "square root", then hence forth "sqr" will be rendered as "sqrt":

Let z = sqrt(3 + 4i)

then z² = 3 + 4i

We want the values of a and b, where a,b are REAL numbers and

z = a + bi

Square both sides to get:

z² = (a + bi)²

= a² + 2abi - b²

= (a² - b²) + 2abi

Thus we have:

3 = a² - b²

4 = 2ab

Solving for b in terms of a, we can recast the last equation as

2 = ab, or b = 2 / a

Substitution leads to

3 = a² - b²

= a² - (2 / a)²

= a² - 4 / a²

Multiply both sides by a² leads to

3a² = a⁴ - 4, or

0 = a⁴ - 3a² - 4

0 = (a² - 4)(a² + 1)

Thus a² = 4 or a² = -1

No real numbers satisfy a² = -1, so all we have is

a = 2 or -2.

In which case,

b = 2 / a

= 1 or -1

So now we have four candidate solutions,

z = 2 + i

z = 2 - i

z = -2 + i

z = -2 - i

Squaring each solution leads 3 + 4i or 3 - 4i. The solutions that when squared lead to 3 + 4i are

z = 2 + i

z = -2 - i