Question

1. # Find The Product Of Z1 And Z2, Where Z1 = 7(Cos 40° + I Sin 40°) And Z2 = 6(Cos 145° + I Sin 145°)

Finding the product of two complex numbers can be a tricky task, but it’s actually not as difficult as it might seem. In this blog post, we’ll explain how to find the product of two complex numbers, z1 and z2. We’ll give a concrete example using two arbitrary complex numbers and then walk through the process step-by-step. By the end, you’ll have all the tools you need to solve similar problems in no time!

## What is the product of z1 and z2?

In mathematics, the product of two complex numbers is calculated using the following formula:

z1 * z2 = (cos θ1 + i sin θ1) * (cos θ2 + i sin θ2)

= cos(θ1 + θ2) + i sin(θ1 + θ2)

where z1 = (cos θ1 + i sin θ1) and z2 = (cos θ2 + i sin θ2).

## How to find the product of z1 and z2

To find the product of z1 and z2, we need to use the polar form of multiplication. In the polar form, the product of two complex numbers is given by:

z1 * z2 = (r1 * r2) * (cos(θ1 + θ2) + i sin(θ1 + θ2))

where r1 and r2 are the magnitudes of z1 and z2, and θ1 and θ2 are their respective angles.

In our case, we have:

z1 = (cos ° + i sin °)
z2 = (cos ° + i sin °)

Therefore, their product is:

z1 * z2 = (|z1| * |z2|) * (cos(arg(z1) + arg(z2)) + i sin(arg(z1) + arg(z2)))
= ((cos ° + i sin °)(cos ° – i sin °))* (cos((-/+)°)+i sin((-/+)°)) //De Moivre’s Formula was used to expand z12 and z22. The (-/+) in front of the angle denotes that the angle could be either positive or negative depending on which way you turn from cosine to sine.

= 1*(cos(+/- 2°)+i sin(+/- 2°))

Therefore, the product of z1 and z2 is (cos(+/- 2°)+i sin(+/- 2°)).

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## What is the best way to calculate the product of z1 and z2?

To calculate the product of z1 and z2, we need to use the definition of multiplication for complex numbers. That is, for two complex numbers a+bi and c+di, their product is (ac-bd)+(ad+bc)i. Applying this to our case, we have:

z1z2 = [(cos ° + i sin °)(cos ° + i sin °)]
= (cos2° – sin2°) + i(2 cos° sin°)
= (1 – 2sin2°) + i(2 cos° sin°)

## Conclusion

In this article, we have discussed how to find the product of two complex numbers. We took an example with z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°). After using the formulae for multiplying complex numbers, we got our final answer which was -44.89 (Cos 185° + I Sin 185°). With this knowledge in hand, you can easily compute products of two or more complex numbers no matter what their values are!

2. Are you ready to tackle a complex math problem? Let’s dive in and see if we can find the product of z1 and z2!

To start, we need to identify our two numbers, z1 and z2. z1 is equal to 7(cos 40° + i sin 40°), and z2 is equal to 6(cos 145° + i sin 145°).

Now that we know our two numbers, we can begin to calculate their product. We can break down our calculations into two steps. First, we will find the real part of the product, and then we will find the imaginary part.

Let’s start with the real part. To calculate this, we will use the formula (a1cosθ1 + a2cosθ2) (b1cosθ1 + b2cosθ2).

Plugging in our numbers, this becomes (7cos40° + 6cos145°) x (7cos40° + 6cos145°).

Simplifying this, we find that the real part of the product is 441.

Now for the imaginary part. To calculate this, we will use the formula (a1cosθ1 – a2cosθ2) (b1sinθ1 + b2sinθ2).

Plugging in our numbers, this becomes (7cos40° – 6cos145°) x (7sin40° + 6sin145°).

Simplifying this, we find that the imaginary part of the product is -109.

Finally, we can combine the real and imaginary parts to get our full product. The product of z1 and z2 is 441 – 109i.

There you have it – the product of z1 and z2! It may have been a bit tricky, but with a bit of patience and the right formula, you can solve this math problem with ease.