Question

If \( z_{1}=x+iy \), \( z_{2}=a+ib \) and \( x^{2}+y^{2}=a^{2}+b^{2} \), then \( z_{2} = \)

• A. \( |z_{1}|\text{cis}(\tan^{-1}(\frac{b}{a})) \)
• B. \( |\overline{z_{2}}|z_{1} \)
• C. \( z_{1}\text{cis}(\tan^{-1}(\frac{y}{x})) \)
• D. \( z_{1} \)

Hint:

Euler's and polar forms are highly efficient tools for complex geometry problems. Remember that \( \text{cis}(\theta) = e^{i\theta} \), and any complex number is explicitly specified by its absolute scale (modulus) and angular direction (argument).

Answer 1

The Correct Option is A

Step 1: Recall polar form.
Any complex number can be written as $z = |z|\,\text{cis}(\phi)$, where $|z|$ is its size and $\phi$ is its angle. Here $\text{cis}(\phi)$ means $\cos\phi + i\sin\phi$.
Step 2: Find the size of each number.
For $z_1 = x+iy$ the size is $|z_1| = \sqrt{x^2+y^2}$. For $z_2 = a+ib$ the size is $|z_2| = \sqrt{a^2+b^2}$.
Step 3: Use the given condition.
We are told $x^2+y^2 = a^2+b^2$. Taking square roots gives \[ |z_1| = |z_2| \] So both numbers have the same size.
Step 4: Find the angle of z2.
The angle of $z_2 = a+ib$ is $\tan^{-1}\!\left(\dfrac{b}{a}\right)$.
Step 5: Write z2 in polar form.
So $z_2 = |z_2|\,\text{cis}\!\left(\tan^{-1}\tfrac{b}{a}\right)$.
Step 6: Swap in the equal size.
Since $|z_2| = |z_1|$, replace it to get \[ \boxed{z_2 = |z_1|\,\text{cis}\!\left(\tan^{-1}\tfrac{b}{a}\right)} \]