Complex Numbers

$i^{403}=$

If $4x+i(3x-y)=3+i(-6)$, where $x$ and $y$ are real numbers, then find the values of $x$ and $y$ respectively.

The value of $\left(\dfrac{-1+i \sqrt{3}}{2}\right)^{18}+\left(\dfrac{-1-i \sqrt{3}}{2}\right)^{18}$ is

The difference between the roots of the equation $x^{2}+2 x+4=0$ is ⋯ (where $i=\sqrt{-1}$ )

If the complex number $z=x+i y$, where $i=\sqrt{-1}$, satisfies the condition $|z+1|=1$, then $z$ lies on

If $1-i$ is a root of the equation $x^{2}+ax+b=0$, where $a,b\in R$, then find the values of $a$ and $b$.

Convert the complex number $\sqrt{3}+i$ in polar form.

The complex number $z=\dfrac{(\cos 2\theta+i\sin 2\theta)^{7}}{(\cos 4\theta+i\sin 4\theta)^{3}}$ in the form $x+iy$ is form $x+iy$ is

The equation of the line joining the points represented by $2-3i$ and $-3+4i$ in the Argand plane is

The value of ${\left( 1+2\omega+\omega^{2} \right)}^{3n}-{\left( 1+\omega+2\omega^{2} \right)}^{3n}$ is :

If $w$ is a complex cube root of unity, then the value of $w^{10}-w^{7}+w^{5}-w^{2}+1$ is.

The square root of (-5-12i) is

$\dfrac{(\cos\theta+i\sin\theta)^{5}(\cos 3\theta-i\sin 3\theta)^{6}}{(\cos 2\theta+i\sin 2\theta)^{3}(\cos 4\theta-i\sin 4\theta)^{5}}=$

If $1,\omega,\omega^{2}$ are cube roots of unity, then ${\left( 1-\omega+\omega^{2} \right)}^{5}+$ ${\left( 1+\omega-\omega^{2} \right)}^{5}$ is