Find the Product of Complex Numbers - Pre-Calculus

Card 0 of 32

Question

Find the value of ${}(1+i)^{100}$ ,where the complex number is given by ${}i^2=-1$ .

Answer

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

Compare your answer with the correct one above

Question

Compute the following sum:

${}1+i+i^2+i^3....+i^{2015}$ . Remember is the complex number satisfying ${}i^{2}=-1$ .

Answer

Note that this is a geometric series.

Therefore we have:

${}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$

Note that,

${}i^{2016}$ = ${}({i}^{2})^{1008}$ and since ${}i^2=-1$ we have ${}{-1}^{1008}=1$ .

$\frac{1-1}{1-i}=\frac{0}{1-i}=0$

this shows that the sum is 0.

Compare your answer with the correct one above

Question

Find the following product.

${}(1-i)^{n}(1+i)^n$

Answer

Note that by FOILing the two binomials we get the following:

${}(1+i)(1-i)=1-i+i-i^2=2$

Therefore,

${}(1-i)^{n}(1-i)^{n}=2^{n}$

Compare your answer with the correct one above

Question

Compute the magnitude of ${}(1+i)^{n}$ .

Answer

We have

${}|(1+i)|=\sqrt{2}$ .

We know that ${}|z^n|=|z|^n$

Thus this gives us,

${}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ .

Compare your answer with the correct one above

Question

Evaluate:

$(2+i)\cdot(1+4i)$

Answer

To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

Compare your answer with the correct one above

Question

Find the product , if

.

Answer

To find the product , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.

Using this method we get the following,

and because

.

Compare your answer with the correct one above

Question

Simplify: $i^4 \times i^3$

Answer

The expression $i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ , the value of .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is:

Compare your answer with the correct one above

Question

Find the product of the two complex numbers

$\small 3+7i$ and $\small 7+10i$

Answer

The product is

$\small (3+7i)(7+10i)=21+30i+49i+70i^2=21-70+79i=-49+79i$

Compare your answer with the correct one above

Question

Find the value of ${}(1+i)^{100}$ ,where the complex number is given by ${}i^2=-1$ .

Answer

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

Compare your answer with the correct one above

Question

Compute the following sum:

${}1+i+i^2+i^3....+i^{2015}$ . Remember is the complex number satisfying ${}i^{2}=-1$ .

Answer

Note that this is a geometric series.

Therefore we have:

${}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$

Note that,

${}i^{2016}$ = ${}({i}^{2})^{1008}$ and since ${}i^2=-1$ we have ${}{-1}^{1008}=1$ .

$\frac{1-1}{1-i}=\frac{0}{1-i}=0$

this shows that the sum is 0.

Compare your answer with the correct one above

Question

Find the following product.

${}(1-i)^{n}(1+i)^n$

Answer

Note that by FOILing the two binomials we get the following:

${}(1+i)(1-i)=1-i+i-i^2=2$

Therefore,

${}(1-i)^{n}(1-i)^{n}=2^{n}$

Compare your answer with the correct one above

Question

Compute the magnitude of ${}(1+i)^{n}$ .

Answer

We have

${}|(1+i)|=\sqrt{2}$ .

We know that ${}|z^n|=|z|^n$

Thus this gives us,

${}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ .

Compare your answer with the correct one above

Question

Evaluate:

$(2+i)\cdot(1+4i)$

Answer

To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

Compare your answer with the correct one above

Question

Find the product , if

.

Answer

To find the product , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.

Using this method we get the following,

and because

.

Compare your answer with the correct one above

Question

Simplify: $i^4 \times i^3$

Answer

The expression $i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ , the value of .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is:

Compare your answer with the correct one above

Question

Find the product of the two complex numbers

$\small 3+7i$ and $\small 7+10i$

Answer

The product is

$\small (3+7i)(7+10i)=21+30i+49i+70i^2=21-70+79i=-49+79i$

Compare your answer with the correct one above

Question

Find the value of ${}(1+i)^{100}$ ,where the complex number is given by ${}i^2=-1$ .

Answer

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

Compare your answer with the correct one above

Question

Compute the following sum:

${}1+i+i^2+i^3....+i^{2015}$ . Remember is the complex number satisfying ${}i^{2}=-1$ .

Answer

Note that this is a geometric series.

Therefore we have:

${}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$

Note that,

${}i^{2016}$ = ${}({i}^{2})^{1008}$ and since ${}i^2=-1$ we have ${}{-1}^{1008}=1$ .

$\frac{1-1}{1-i}=\frac{0}{1-i}=0$

this shows that the sum is 0.

Compare your answer with the correct one above

Question

Find the following product.

${}(1-i)^{n}(1+i)^n$

Answer

Note that by FOILing the two binomials we get the following:

${}(1+i)(1-i)=1-i+i-i^2=2$

Therefore,

${}(1-i)^{n}(1-i)^{n}=2^{n}$

Compare your answer with the correct one above

Question

Compute the magnitude of ${}(1+i)^{n}$ .

Answer

We have

${}|(1+i)|=\sqrt{2}$ .

We know that ${}|z^n|=|z|^n$

Thus this gives us,

${}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ .

Compare your answer with the correct one above

Tap the card to reveal the answer