262hint for 1 and 2 - put z=x+iy and then compare the real and imaginary parts
hint for 3 - rationalise the denominator and use the given condition.
hint for 4- the expression can be written as (z2+1)(z2+z+1)=0
1. find the number of complex numbers satisfying |z| = z + 1 + 2i.
2. If z + √2 |z + 1| + i = 0 and z = x + iy then x = ?
3. if p2 + q2 = 1, p,q E R, then 1 + p + iq1 + p - iq is equal to
(A) p + qi
(B) p - qi
(C) q + pi
(D) q - pi
4. if z be a complex number satisfying z4 + z3 + 2z2 + z + 1 = 0, then find the value of |z|
Answers.
1. 1
2. -2
3. A
4. 1
-
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8 Answers
262
3well i have posted all the solutions on goiit.
you asked the same question there too.
341quicker way for 3:
Remember that if |z| =1 we have
\overline{z} = \frac{1}{z}
So the given expression is
\frac{1+z}{1+\overline{z}} = \frac{1+z}{1+\frac{1}{z}}=z
36one request...
can someone post the solution to the second question please?
1708\hspace{-16}(2)\;z+\sqrt{2}.\mid z+1\mid +i=0\\\\ $\underbrace{\sqrt{2}.\mid z+1\mid}_{\mathbf{L.H.S}}=\underbrace{-z-i}_{\mathbf{R.H.S}}$\\\\ $So $\mathbf{L.H.S}$ is a Real no. So $\mathbf{R.H.S}$ is also a real no.\\\\ So $z=x-i\;,$ Where $x<0$\\\\ So $\sqrt{2}.\mid (x+1)-i\mid=-x$\\\\ $(x+1)^2+1=\frac{x^2}{2}$\\\\ $x^2+4x+4=0\Leftrightarrow (x+2)^2=0$\\\\ So $x=-2$\\\\ So $\boxed{\boxed{x=-2-i}}$