Find the root of the complex 3+4i

What we should know is that the root 3+4i is a complex number that looks alot like a+bi.

We can say : rt(3+4i) = a+bi (Where we dont know what a & b is..yet)

and when we square both sides (rt(3+4i))^2=(a+bi)^2 | 3+4i = (a+bi)^2

we get 3+4i = a^2+2abi-b^2

We seperate the Real and Imaginary parts to get a simultainus equation

3 = a^2-b^2

4 = 2ba

if this is solved we get a= (+-)2 and b =(+-)1

to get (+-)(2+i) <--- which is the answer

AA
Answered by Ade A. Further Mathematics tutor

3230 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Solve the equation 3sinh(2x) = 13 - 3e^(2x), answering in the form 0.5ln(k). where k is an integer


How do I differentiate tan(x) ?


Integrate f(x) = 1/(1-x^2)


How do you deal with 3 simultaneous equations? (Struggling with Q7 of AQA specimen paper 1)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning