Complex Numbers Problem

[Soup674]

Question:
Given that (a + bi)^2 = 48 + 14i obtain a pair of simultaneous equations
involving a and b. Hence find the two square roots of 48 + 14i.

What I've done so far:
[math]a^2+2abi+b^2=48+14i \\\therefore a^2+b^2=48 \\2abi=14i[/math]

 

[topsquark]

Good, except for one little point:
[imath](a + bi)^2 = a^2 + 2abi + i^2b^2 = a^2 + 2abi - b^2[/imath]

What I'd do from here is solve [imath]2ab = 14[/imath] for b in terms of a and plug that into the [imath]a^2 - b^2 = 48[/imath] equation to get a.

-Dan

 

[Soup674]

Good, except for one little point:
[imath](a + bi)^2 = a^2 + 2abi + i^2b^2 = a^2 + 2abi - b^2[/imath]

What I'd do from here is solve [imath]2ab = 14[/imath] for b in terms of a and plug that into the [imath]a^2 - b^2 = 48[/imath] equation to get a.

-Dan

Ah! Thank you so much! Here's the answer for anyone who comes looking:
[math]b={7\over a} \\a^2-{49\over a^2}=48 \\a^4-48a^2-49=0 \\a^4+a^2-49a^2-49=0 \\a^2(a^2+1)-49(a^2+1)=0 \\(a^2-49)(a^2+1)=0 \\a=\pm7\thickspace(a∈\R) \\b=\pm1 \\\therefore \sqrt{49+14i}=\pm(7+i)[/math]

 

[Cubist]

Well done @Soup674

There's a very minor typo on the last line...
[math] \sqrt{\cancel{49}\, 48+14i}=\pm(7+i) [/math]