Given two complex numbers, $z_1 = 4 + 3i$ and $z_2 = 2 – 5i$, the real part of $(z_1 \bar{z_2})$ is ________.
Understanding the Problem
We are given two complex numbers, $z_1 = 4 + 3i$ and $z_2 = 2 – 5i$. The task is to find the real part of the product $z_1 \bar{z_2}$, where $\bar{z_2}$ is the complex conjugate of $z_2$. The provided answer suggests the value is -7.
Calculating the Conjugate
First, find the complex conjugate of $z_2$. If $z_2 = a + bi$, its conjugate $\bar{z_2} = a - bi$.
Given $z_2 = 2 - 5i$, the conjugate is:
$ \bar{z_2} = 2 - (-5i) = 2 + 5i $
Calculating the Product $z_1 \bar{z_2}$
Next, multiply $z_1$ by $\bar{z_2}$: $ z_1 \bar{z_2} = (4 + 3i)(2 + 5i) $ Use the distributive property (FOIL method): $ = 4(2) + 4(5i) + 3i(2) + 3i(5i) $ $ = 8 + 20i + 6i + 15i^2 $ Since $i^2 = -1$: $ = 8 + 20i + 6i + 15(-1) $ $ = 8 + 20i + 6i - 15 $ Combine the real terms ($8$ and $-15$) and the imaginary terms ($20i$ and $6i$): $ = (8 - 15) + (20 + 6)i $ $ = -7 + 26i $
Identifying the Real Part
The result of the product $z_1 \bar{z_2}$ is the complex number $-7 + 26i$. The real part of a complex number $a + bi$ is $a$. Therefore, the real part of $-7 + 26i$ is $-7$.
Important Questions from Complex Numbers
- Consider the following complex numbers
$$ z_1 = r_1(\cos \theta_1 + i \, \sin \theta_1) $$
$$ z_2 = r_2(\cos \theta_2 + i \, \sin \theta_2) $$
where $r_1, r_2$ are real numbers, $0 \leq \theta_1 \leq \frac{\pi}{2}$, $0 \leq \theta_2 \leq \frac{\pi}{2}$, and $i = \sqrt{-1}$
If $|z_1 + z_2| = |z_1| + |z_2|$, which one of the following conditions is necessarily CORRECT? - What is the value of \((2 - 3i) + (3 + 5i)\)
- What is the real part of the number \(15 + 2i\)
- What is the real part of the number $5i$?
- What is the value of $(6 + 3i) + (2 + 5i)$?
