Question

Find conjugate of √-3 + 4i^2

Answer 1

We have to find the conjugate of $\sf{\sqrt{-3}+4i^2.}$

The conjugate of a complex number $\sf{z=a+ib}$ is given by,

$\longrightarrow\sf{\bar z=a-ib}$

for $\sf{a,\ b\in\mathbb{R}.}$

But here,

$\longrightarrow\sf{z=\sqrt{-3}+4i^2}$

We have to make $\sf{z}$ in $\sf{a+ib}$ form.

We know that,

• $\sf{i=\sqrt{-1}}$

• $\sf{i^2=-1}$

Then,

$\longrightarrow\sf{z=\sqrt{-1}\times\sqrt3+4\times-1}$

$\longrightarrow\sf{z=i\sqrt3-4}$

$\longrightarrow\sf{z=-4+i\sqrt3}$

Therefore,

• $\sf{a=-4}$

• $\sf{b=\sqrt3}$

Hence the conjugate is,

$\longrightarrow\sf{\underline{\underline{\bar z=-4-i\sqrt3}}}$

Answer 2

Answer:

here , √-3 + 4i² = √3 x √i + 4 x (-1). ( i² = -1)

= √3i - 4= -4 + √3i

so, the conjugate of

-4 + √3i = -4 - √3i.