Find the square root of complex number 3 + 4i

Strategy: write down an equation satisfied by the square root, and solve it algebraically.  Method:  square root x+iy  satisfies (x+iy)2 = 3 + 4i. Expand: x2-y2 +2xyi = 3+4i. Comparing coefficients gives:   x2-y2 =3 and 2xy =4. Then substitute y:  x2 -4/x2 = 3. Rearrange to get quadratic in x2 : (x+1)(x2 -4) = 0. x can't be imaginary (by definition) so x= +/- 2. Plug in to equation 2xy = 4, get y = +/- 1. So square root is +/- (2+i).

JS
Answered by Jakob S. Further Mathematics tutor

28864 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Expand (1+x)^3. Express (1+i)^3 in the form a+bi. Hence, or otherwise, verify that x = 1+i satisfies the equation: x^3+2*x-4i = 0.


I'm struggling with an FP2 First-Order Differential Equations Question (Edexcel June 2009 Q3) and the topic in general!


Let I(n) = integral from 1 to e of (ln(x)^n)/(x^2) dx where n is a natural number. Firstly find I(0). Show that I(n) = -(1/e) + n*I(n-1). Using this formula find I(1).


By Differentiating from first principles, find the gradient of the curve f(x) = x^2 at the point where x = 2


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:

© 2026 by IXL Learning