We expand the term (a+ib)³ using the binomial theorem (A+B)³ = A³ + 3A²B + 3AB² + B³. Remember that i² = -1 and i³ = -i. We then group the real and imaginary parts.

If $(x+i y)^{\frac{1}{3}}=a+i b$ , where $x, y, a, b \in \mathrm{R}$ , show that $\frac{x}{a}-\frac{y}{b}=-2\left(a^{2}+b^{2}\right)$

Answer: The identity $\frac{x}{a}-\frac{y}{b}=-2\left(a^{2}+b^{2}\right)$ is shown to be true.

Step-by-step solution

Step 1: Cube both sides of the equation

To eliminate the fractional exponent, we cube both sides of the given equation. This allows us to express $x+iy$ in terms of $a$ and $b$ .

Step 2: Expand the right side using the binomial theorem

We expand the term $(a+ib)^3$ using the binomial theorem $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$ . Remember that $i^2 = -1$ and $i^3 = -i$ . We then group the real and imaginary parts.

Step 3: Equate real and imaginary parts

Since $x+iy = (a^3-3ab^2) + i(3a^2b-b^3)$ , and $x, y, a, b$ are real numbers, we can equate the real parts and the imaginary parts from both sides of the equation.

Step 4: Simplify the terms $\frac{x}{a}$ and $\frac{y}{b}$

We divide the expression for $x$ by $a$ and the expression for $y$ by $b$ . We can factor out $a$ from the expression for $x$ and $b$ from the expression for $y$ before dividing, assuming $a \neq 0$ and $b \neq 0$ .

Step 5: Substitute and simplify to show the required identity

Finally, we substitute the simplified expressions for $\frac{x}{a}$ and $\frac{y}{b}$ into the left side of the identity we need to prove. Combining like terms leads directly to the right side of the identity.

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If (x+iy)13=a+ib(x+i y)^{\frac{1}{3}}=a+i b(x+iy)31​=a+ib, where x,y,a,b∈Rx, y, a, b \in \mathrm{R}x,y,a,b∈R, show that xa−yb=−2(a2+b2)\frac{x}{a}-\frac{y}{b}=-2\left(a^{2}+b^{2}\right)ax​−by​=−2(a2+b2)