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# Find in a + bi form?

(5-7i)^.5

Any tricks to this I tried Demoivre theorem but it gets too messy.

### 3 Answers

- trueproberLv 76 years agoFavorite Answer
Hello Jack P, let us assume (5 -7i)^1/2 = a + i b

Squaring on both sides we have 5 -7i = (a+ib)^2

Or a^2 - b^2 + i (2 ab) = 5 - 7i

Equating real and imaginary parts on both sides you have

a^2 - b^2 = 5 and 2 ab = -7

Now( a^2 + b^2 )^2 = (a^2 - b^2)^2 + 4 (ab)^2

So ( a^2 + b^2 )^2 = 25 + 196 = 221

( a^2 + b^2 )^2 = 221

Or ( a^2 + b^2 ) = ./221

Along with ( a^2 - b^2 ) = 5 you can solve for a^2 and b^2 where from you can get a and b

Hence in a+ib form.

- Anonymous6 years ago
It's not going to be nice.

The modulus of z = a+bi is √a^2+b^2

The absolute square of z is |z|^2 = a^2 + b^2

The modulus of complex numbers z and w obeys the rule |z| times |w| = |zw|

Same is true of the absolute square, obviously.

Now let z = 5 - 7i

The absolute square of z = 25+49 = 74

The let w = a+bi = √z

The absolute square of w = a^2 + b^2 = √74

That tells you that a or b or both must be irrational. They aren't going to be nice numbers

If the absolute square is a perfect square, then the square root will have nicer terms. E.g., 4+3i has modulus squared of 25, and √4+3i = √4.5 + (√2/2). So a and b will be square roots of a rational number. Here they involve square roots of irrational numbers.

I used Excel to calculate w = ±(2.60790387735463-1.34207400448757i)

2.6079 = √((+5 + √(25+4*12.25))/2)

1.34207 =√((-5 + √(25+4*12.25))/2)

- EliotLv 56 years ago
= (√74, arctan(−⁷⁄₅))⁵

= (√74⁵, 5arctan(−⁷⁄₅))

Now get out your calculator and do the rest.