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.