Problem 21

Question

$$(1+i)(10+10 i)=10(1+i)^{2}=20 i$$

Step-by-Step Solution

Verified

Answer

The equation is correct: both sides simplify to $20i$.

1Step 1: Simplify the Left Expression

First, we need to multiply the complex numbers on the left side, $(1+i)(10+10i)$. Use the distributive property to expand this: $(1+i)(10+10i) = 1 \cdot 10 + 1 \cdot 10i + i \cdot 10 + i \cdot 10i$. Simplifying this expression gives $10 + 10i + 10i + 10i^2$. Since $i^2 = -1$, we have $10 + 20i - 10 = 20i$.

2Step 2: Simplify the Right Expression

Now let's verify the right side, $10(1+i)^2$. First, calculate $(1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i$. Then, multiply by 10: $10 \cdot 2i = 20i$.

3Step 3: Compare Both Sides

Now, we compare the simplified left expression and the right expression. Both resulted in $20i$. As they are equal and the simplifications match, it confirms the equation $(1+i)(10+10i) = 10(1+i)^2 = 20i$ is correct.

Key Concepts

Distributive PropertySimplificationImaginary Unit i

Distributive Property

The Distributive Property is an essential tool when working with complex numbers. It allows us to simplify expressions by multiplying terms individually. For instance, when you deal with the product

$(1+i)(10+10i)$,

you apply the distributive property to each term within the parentheses.
Here's how it works:

First, multiply $1$ by each term in the second parentheses: $1 \cdot 10 + 1 \cdot 10i$.
Then multiply $i$ by each of those terms: $i \cdot 10 + i \cdot 10i$.

Add all these together to get $1 \cdot 10 + 1 \cdot 10i + i \cdot 10 + i \cdot 10i$,
leading to $10 + 10i + 10i + 10i^2$. Understand how each term arises helps you master this property.

Simplification

Simplifying complex expressions is important to reach the correct solution. In our example and many similar tasks, simplification leads to the final answer. After applying the distributive property, you end up with

$10 + 20i + 10i^2$.

Simultaneously, remember that

$i^2 = -1$, making

$10i^2 = -10$.
As a result, the expression simplifies further to $10 + 20i - 10$.
Finally, $10 - 10$ becomes zero, leaving you with $20i$.
Always check each step for opportunities to simplify, either through combining like terms or replacing known values like

$i^2$

with $-1$.

Imaginary Unit i

The imaginary unit, represented as

$i$,

is the cornerstone of complex numbers. It's defined such that

$i^2 = -1$.

Understanding this allows us to explore numbers beyond the real number line.
When handling complex expressions, knowing the impact of $i$
and its powers is vital. For example, identifying $i^2 = -1$
enables you to simplify terms like $10i^2$ to $-10$. This crucial property helps transform expressions with higher powers of $i$
into more manageable forms.

Problem 21

Problem 22

Other exercises in this chapter

Problem 21

$$u=e^{x} \cos y, \quad v=-e^{x} \sin y ; \quad \frac{\partial u}{\partial x}=e^{x} \cos y, \quad \frac{\partial v}{\partial y}=-e^{x} \cos y ; \quad \frac{\par

View solution

Problem 21

$$(2 \sqrt{2})^{5}\left[\cos \left(-\frac{5 \pi}{4}\right)+i \sin \left(-\frac{5 \pi}{4}\right)\right]=-128+128 i$$

View solution

Problem 22

$\sin z=i \sinh 2$ implies $\sin x \cosh y+i \cos x \sinh y=0+i \sinh 2$ and so we must have $\sin x \cosh y=0$ and $\cos x \sinh y=\sinh 2 .$ since \(\

View solution

Problem 22

$$\lim _{z \rightarrow 1+i} \frac{z^{2}-2 z+2}{z^{2}-2 i}=\lim _{z \rightarrow 1+i} \frac{[z-(1+i)][z-(1-i)]}{[z-(1+i)][z-(-1-i)]}=\frac{1}{2}+\frac{1}{2} i$$

View solution