# What is the complex number that has the square of 5+12i ?

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### 2 Answers

Let the number that has a square of 5 + 12i be a + bi

(a + bi)^2 = 5 + 12i

use (a + b)^2 = a^2 + b^2 + 2*a*b

=> a^2 + b^2*i^2 + 2*a*b*i = 5 + 12i

i^2 = -1

=> a^2 - b^2 + 2*a*b*i = 5 + 12i

equate the real and complex coefficients

=> a^2 - b^2 = 5 and ab = 6

a = 6/b

substitute in a^2 - b^2 = 5

=> 36/b^2 - b^2 = 5

=> 36 - b^4 = 5b^2

=> b^4 + 5b^2 - 36 = 0

=> b^4 + 9b^2 - 4b^2 - 36 = 0

=> b^2( b^2 + 9) - 4(b^2 + 9) = 0

=> (b^2 - 4)(b^2 + 9) = 0

=> b^2 = 4 and b^2 = -9

b is a real number, so we eliminate b^2 = -9

b^2 = 4

=> b = 2 and b = -2

a = 3 and a = -3

**The required number can be 3 + 2i and -3 -2i**

We'll write the rectangular form of the complex number that has to be found.

z = a + b*i

We'll raise to square:

z^2 = a^2 + 2ab*i + b*i^2, where i^2 = -1

z^2 = a^2 - b^2 + 2ab*i (1)

We know, from enunciation, that the square of the complex number is 5 + 12i (2).

We'll equate (1) = (2):

a^2 - b^2 + 2ab*i = 5 + 12i

Comparing, we'll get:

a^2 - b^2 = 5 (3)

2ab = 12

ab = 6

b = 6/a (4)

We'll solve the system:

a^2 - 36/a^2 = 5

a^4 - 36 - 5a^2 = 0

We'll solve bi-quadratic equation. Let t^2 = a^4

t^2 - 5t - 36 = 0

t1 = [5+sqrt(25+144)]/2

t1 = (5+13)/2

t1 = 9

t2 = -4

t1 = a^2 => a^2 = 9 => a1 = 3 and a2 = -3

b1 = 6/a1 = 6/3

b1 = 2 and b2 = -2

a^2 = t2 => a^2 = -4

a3 = 2i and a4 = -2i

b1 = 6/2i = -12i/4 = -3i

b2 = 6/-2i = 3i

**The complex numbers whose square is 5 + 12i are: z1 = 3 + 2i and z2 = -3 - 2i.**