Learn Complex Numbers and Quadratic Equations with Free Lessons & Tips

Certainly! Let's simplify the given expression step by step:

Given complex numbers: (1−i)−(−1+i6)(1−i)−(−1+i6)

First, let's distribute the negative sign: (1−i)+(1−i6)(1−i)+(1−i6)

Now, let's combine like terms: 1+1−i−i61+1−i−i6

Combine the real parts (1 + 1 = 2) and the imaginary parts (-i - i6 = -7i): 2−7i2−7i

So, the given complex number (1−i)−(−1+i6)(1−i)−(−1+i6) in the form a+iba+ib is 2−7i2−7i.

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Sure! When expressing a complex number in polar form, we represent it as r(cos⁡θ+isin⁡θ)r(cosθ+isinθ), where rr is the magnitude of the complex number and θθ is the angle it makes with the positive real axis.

For the complex number −3−3, the magnitude is ∣r∣=∣−3∣=3∣r∣=∣−3∣=3, and the angle it makes with the positive real axis is ππ (or 180 degrees) in the clockwise direction.

So, in polar form, −3−3 can be expressed as 3(cos⁡π+isin⁡π)3(cosπ+isinπ) on UrbanPro.

As the discriminant (b2−4ac)(b2−4ac) is negative, the roots will be complex.

So, the solutions for the quadratic equation 2x2+x+1=02x2+x+1=0 are:

x=−1+i74x=4−1+i7 x=−1−i74x=4−1−i7

Where ii is the imaginary unit.

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When dealing with complex numbers, it's essential to grasp the operations involving both their real and imaginary parts.

Let's take two complex numbers, z1=a+biz1=a+bi and z2=c+diz2=c+di, where aa, bb, cc, and dd are real numbers and ii represents the imaginary unit.

Now, to find the real part of the product z1z2z1z2, we perform the multiplication:

z1z2=(a+bi)(c+di)z1z2=(a+bi)(c+di)

Expanding this expression:

z1z2=ac+adi+bci+bdi2z1z2=ac+adi+bci+bdi2

Remembering that i2=−1i2=−1, we simplify:

z1z2=ac+adi+bci−bdz1z2=ac+adi+bci−bd

Now, let's separate the real and imaginary parts:

Re(z1z2)=ac−bdRe(z1z2)=ac−bd

This is indeed the real part of the product of z1z1 and z2z2. Now, let's look at the right side of the equation:

Re(z1)=aRe(z1)=a Re(z2)=cRe(z2)=c

Im(z1)=bIm(z1)=b Im(z2)=dIm(z2)=d

Now, substituting these into the expression Re(z1)Re(z2)−Im(z1)Im(z2)Re(z1)Re(z2)−Im(z1)Im(z2), we get:

Re(z1)Re(z2)−Im(z1)Im(z2)=ac−bdRe(z1)Re(z2)−Im(z1)Im(z2)=ac−bd

And this is exactly what we got for Re(z1z2)Re(z1z2) earlier.

So, we've shown that Re(z1z2)=Re(z1)Re(z2)−Im(z1)Im(z2)Re(z1z2)=Re(z1)Re(z2)−Im(z1)Im(z2). This is a fundamental property when dealing with complex numbers, and understanding it thoroughly will assist you in various mathematical applications. If you have any further questions or need additional clarification, feel free to ask!

=2

So, the modulus of the given expression is 22. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is always here to help you excel in your studies.

As an experienced tutor registered on UrbanPro, I can guide you through solving this problem step by step. Let's break it down:

We have the expression (x−iy)(3+5i)(x−iy)(3+5i), and we want it to be the conjugate of −6−24i−6−24i.

First, let's multiply the complex numbers:

(x−iy)(3+5i)=3x+5ix−3iy−5y(x−iy)(3+5i)=3x+5ix−3iy−5y

Now, let's rearrange this expression:

=(3x−5y)+i(5x−3y)=(3x−5y)+i(5x−3y)

Now, we want this expression to be the conjugate of −6−24i−6−24i. The conjugate of a complex number a+bia+bi is a−bia−bi.

So, we need to set:

3x−5y=−63x−5y=−6 (for the real parts to match)

5x−3y=−245x−3y=−24 (for the imaginary parts to match)

Now, we have a system of linear equations:

3x−5y=−63x−5y=−6 5x−3y=−245x−3y=−24

We can solve this system using any method we prefer, such as substitution or elimination. Let's use elimination:

Multiply the first equation by 5 and the second equation by 3 to eliminate yy:

15x−25y=−3015x−25y=−30 15x−9y=−7215x−9y=−72

Now, subtract the second equation from the first:

−25y−(−9y)=−30−(−72)−25y−(−9y)=−30−(−72)

−16y=42−16y=42

y=−4216=−218y=−1642=−821

Now, substitute y=−218y=−821 into one of the original equations to solve for xx. Let's use the first equation:

3x−5(−218)=−63x−5(−821)=−6

3x+1058=−63x+8105=−6

3x=−6−1058=−488−1058=−15383x=−6−8105=−848−8105=−8153

x=−15324=−178x=−24153=−817

So, the real numbers xx and yy that satisfy the given conditions are x=−178x=−817 and y=−218y=−821.

As an UrbanPro tutor, I recommend practicing similar problems to strengthen your understanding of complex numbers and their operations. Feel free to ask if you have any further questions!

2. • y+1=0.

   Solving these equations will give us possible values for yy.

By finding solutions for xx and yy, we'll get complex numbers satisfying the given equation. If you need further clarification or assistance, feel free to reach out. Remember, UrbanPro is your gateway to top-notch tutoring assistance!