The Modulus and the Conjugate of a Complex Number

Find the conjugate of $(6+5\mathrm{i}{)}^2$.

$\left|\frac{\left(3+i\right)\left(2-i\right)}{1+i}\right|=$

$f\left(x\right)=7x^5+3x^4+a_1{x}^3+a_2{x}^2+a_{3}x+1$ such that for any $z=x+iy$ where $xy\ne 0$ and $x, y\in R$, $f\left(\overline{z}\right)=\overline{f\left(z\right)}$

then the product of $a_1, a_2, a_3$ is

If $z=x+iy$ and $x^2+y^2=16$, then the range of $\left|\left|x\right|-\left|y\right|\right|$ is

Find the real part of the complex number ${\left(1-2i\right)}^{-3}$

If $z$ is a complete number, then $z^2+{\overline{z}}^2=2$ represents

Define $f:C\to R$ by $f\left(z\right)=\left|z\right|\forall z\in C.$ Then which of the following is false?

The modulus of $e^{iz}$, if $z=x+iy$ is a complex number, where $x$ and $y$ are real numbers, is

The modulus amplitude form of complex number $-1-i\sqrt{3}$ is

If $z=\cos \frac{\pi }{4}+i\sin \frac{\pi }{6},$ then

The modulus of the complex number $\frac{(1-i\sqrt{3})(\cos \theta +i\sin \theta )}{(1+i)(\cos \theta -i\sin \theta )}$ is

If $x+iy=(1+i)(2+i)(3+i),$ then $x^2+y^2$ is equal to

For any complex number $z,$ which of the following is not true?

If $|z-i|<|z+i|$, then

Find $\left|7-24i\right|$.

Let $z$ be a complex number such that $\left|z\right|+z=3+i$ , (where $i=\sqrt{-1}$ ). Then $\left|z\right|$ is equal to:

If $\left|z-1\right|=\left|z+1\right|=\left|z-2i\right|$, then the value of $\left|z\right|$ is

The complex number which satisfies the equation $z+\sqrt{2} \left|z+1\right|+i=0$ is (where $i=\sqrt{-1}$)

If $z_1$and $z_2$ are two complex numbers such that $\left|z_1\right|=\left|z_2\right|+\left|z_1-z_2\right|, \left|z_1\right|>\left|z_2\right|,$ then

For $x_1,x_2,y_1,y_2\in R,$ if $0<x_1<x_2 , y_1=y_2$ and $z_1=x_1+i{y}_1 ,z_2=x_2+i{y}_2$ and $z_3=\frac{1}{2}\left(z_1+z_2\right),$ then $z_1, z_2$ and $z_3$ satisfy :