## - Square Root of a Complex Number

LESSON 1

Find $$\sqrt{15+8i}$$

SOLUTION
We assume that the square root of a complex number is a complex number
$$\sqrt{15+8i}=a+bi$$
• Square both sides and simplify
$$15+8i=(a+bi)^2$$
$$15+8i=a^2-b^2+2abi$$
• Equate real parts
$$a^2-b^2=15$$ (1)
• Equate imaginary parts
$$2ab=8$$
$$\displaystyle a={4\over b}$$ (2)
• Substitute (2) into (1) and solve for $$a$$ and $$b$$
$$\displaystyle \left({4\over b}\right)^2-b^2=15$$
$$\displaystyle {16\over b^2} -b^2=15$$
$$b^4+15b^2-16=0$$
$$(b^2+16)(b^2-1)=0$$
$$b^2=-16$$ Invalid since $$b$$ is real
$$b^2=1$$
$$b=±1$$
$$a=±4$$
• Substitute values for a and b into a+bi
$$\sqrt{15+8i}=4+i$$ or $$-4-i$$
$$=±(4+i)$$
Thus, we see that a complex number has 2 square roots, which are complex numbers.