Math Interactive
13. Complex Numbers
13.2. Complex algebra
13.2.3. Comparing parts
Important

To solve more complicated equations, we may need to rely on comparing real and imaginary parts

Equality of complex numbers

Two complex numbers a+bia+b\mathrm{i} and x+yi,x+y\mathrm{i}, where a,b,x,y∈R,a,b,x,y \in \mathbb{R}, are considered equal if and only if both their real parts and imaginary parts are equal.

If a+bi=x+yi,a+b\mathrm{i} = x+y\mathrm{i}, where a,b,x,y∈R,a,b,x,y \in \mathbb{R}, then a=xa=x and b=y.b=y.

Remark: Note that a,b,x,ya,b,x,y above are real. If we have u+vi=w+zi,u+v\mathrm{i} = w + z\mathrm{i}, where u,v,wu,v,w and zz are complex, then the above result does not hold. Do you know why?

Examples

Example 1: square root of a complex number

Step 1: Let z=x+yi{z=x+y\mathrm{i}}

Let z=x+yi,{z=x+y\mathrm{i},} where x{x} and y{y} are real.

(x+yi)2=−3−4ix2−y2+2xyi=−3−4i \begin{aligned} (x+y\mathrm{i})^2 &= - 3 - 4 \mathrm{i} \\ x^2 - y^2 + 2xy \mathrm{i} &= - 3 - 4 \mathrm{i} \end{aligned}
Step 1: Let z=x+yi{z=x+y\mathrm{i}}

Example 2: complex equation with conjugates

Step 1: Let z=x+yi{z=x+y\mathrm{i}}

Let z=x+yi,{z=x+y\mathrm{i},} where x{x} and y{y} are real.

−2x+yi+(x+yi)(x−yi)=3+4i−2x+x2+y2−2yi=3+4i \begin{aligned} - 2 x + y \mathrm{i} + (x+y\mathrm{i})(x-y\mathrm{i}) &= 3 + 4 \mathrm{i} \\ - 2 x + x^2 + y^2 - 2 y\mathrm{i} &= 3 + 4 \mathrm{i} \\ \end{aligned}
Step 1: Let z=x+yi{z=x+y\mathrm{i}}
Next: Conjugate root theorem >>