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2002## Complex Calculations

Find all fourth roots of the complex number -7+24*i*. Express your answers in the form a+b*i*.

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## Complex Calculations Solution

Considering the equation x^{4}=-7+24*i*, it is probably easier to first solve for x^{2}, and then for x. Let a+b*i*=x^{2}

(x^{2})^{2}=(a+b*i*)^{2}=a^{2}-b^{2}+2ab*i*

We know, then, that a^{2}-b^{2} is the real coefficient, -7.

Now, our original complex *vector* has a magnitude of √((a^{2}-b^{2})^{2}+(2ab)^{2})=a^{2}+b^{2}, and also √((-7)^{2}+24^{2})=25

So, a^{2}+b^{2}=25,

and a^{2}-b^{2}=-7

Adding the two equations yields

2a^{2}=18

a^{2}=9

a=3 or a=-3

Now, since 2ab=24, b=4 or b=-4

Basically, the two square roots of -7+24*i* are 3+4*i* and -(3+4*i*), so our fourth roots of -7+24*i* are going to be both square roots of 3+4*i* and *i* multiplied by both square roots of 3+4i.

Following a similar procedure, let the square root of 3+4i equal c+d*i*.

We know that c^{2}-d^{2}=3, and that the magnitude of 3+4i, 5, equals c^{2}+d^{2}

c^{2}-d^{2}=3

c^{2}+d^{2}=5

2c^{2}=8

c=2 or c=-2

2cd=4

d=1 or d=-1

So the square roots of 3+4*i* are 2+*i* and -2-*i*, and the square roots of -(3+4*i*) are -1+2*i* and 1-2*i*.

It's worth noting that, by the principle derived for finding square roots, it must be true that every time a complex number with integer coefficients is squared, a "pythagorean triple" is formed. For example,

(4+*i*)^{2}=15+8*i*

8^{2}+15^{2}=17^{2}

(8+3*i*)^{2}=55+48*i*

48^{2}+55^{2}=73^{2}

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