Question

Given that $i^{2}=-1$. If $z_{1}=(7+i\sqrt{5})^{2}+(7-i\sqrt{5})^{2}$ and $z_{2}=(3+2i)^{3}-(3-2i)^{3}$, then

• A. $z_{1}$ is a purely imaginary number and $z_{2}$ is a purely real number
• B. $z_{1}$ is a purely real number and $z_{2}$ is a purely imaginary number
• C. both $z_{1}$ and $z_{2}$ are purely imaginary numbers
• D. both $z_{1}$ and $z_{2}$ are purely real numbers
• E. $z_{1}+z_{2}$ is a purely real number

Hint:

Math Tip: You do not need to expand large binomials with complex numbers! Using substitution $z = w + \bar{w}$ guarantees a real number, and $z = w - \bar{w}$ guarantees a purely imaginary number.

Answer 1

The Correct Option is B

Concept:
Complex Numbers - Properties of Conjugates.
For any complex number $z$:

• If $z = \bar{z}$, then $z$ is purely real.
• If $z = -\bar{z}$, then $z$ is purely imaginary.

Step 1: Analyze the expression for $z_1$.
Let $w = 7+i\sqrt{5}$. The conjugate of $w$ is $\bar{w} = 7-i\sqrt{5}$.
We can rewrite $z_1$ in terms of $w$: $$ z_1 = w^2 + \bar{w}^2 $$
Step 2: Take the conjugate of $z_1$.
Apply the property that the conjugate of a sum is the sum of conjugates, and the conjugate of a power is the power of the conjugate: $$ \bar{z}_1 = \overline{w^2 + \bar{w}^2} $$ $$ \bar{z}_1 = \overline{w^2} + \overline{\bar{w}^2} $$ $$ \bar{z}_1 = \bar{w}^2 + w^2 $$ Since $\bar{z}_1 = z_1$, the number $z_1$ is a purely real number.
Step 3: Analyze the expression for $z_2$.
Let $u = 3+2i$. The conjugate of $u$ is $\bar{u} = 3-2i$.
We can rewrite $z_2$ in terms of $u$: $$ z_2 = u^3 - \bar{u}^3 $$
Step 4: Take the conjugate of $z_2$.
Apply the same conjugate properties: $$ \bar{z}_2 = \overline{u^3 - \bar{u}^3} $$ $$ \bar{z}_2 = \overline{u^3} - \overline{\bar{u}^3} $$ $$ \bar{z}_2 = \bar{u}^3 - u^3 $$ Factor out a negative sign: $$ \bar{z}_2 = -(u^3 - \bar{u}^3) = -z_2 $$ Since $\bar{z}_2 = -z_2$, the number $z_2$ is a purely imaginary number.