Question

abs val of i*z(conjugate) is equal to 2sqrt2 z=a+i i^2=-1 a=?

Answer 1

i am not getting this problem. is it
\[z=a+i, \overline z = a-i, |i^{\overline z}|=2\sqrt{2}\]?

Answer 2

second one

Answer 3

sorry third

Answer 4

i wrote the whole problem. is that the whole thing?

Answer 5

that is, you are sure that
\[z=a+i\] and so
\[\overline z =a-i\] rigtht

Answer 6

yes its the whole thing

Answer 7

then i am confused. you have
\[i=e^{\frac{\pi}{2}}\]
\[i^{a-i}=e^{\frac{\pi}{2}i(a-i)}\]
\[=e^{\frac{\pi}{2}}e^{\frac{a\pi}{2}i}\] and it looks to me like the absolute value of this thing is
\[e^{\frac{\pi}{2}}\]so maybe i am missing something

Answer 8

typo on first line should read
\[i=e^{\frac{\pi}{2}i}\]

Answer 9

hold on, is this
\[i\times \overline z\] or
\[i^{\overline z}\]?

Answer 10

first one

Answer 11

if it is the first one then no problem. yoju get
\[i(a-i)=1+ai\] so
\[|1+ai|=\sqrt{1^2+a^2}=2\sqrt{2}\] and so
\[1+a^2=8\] making
\[a^2=7\] and thus
\[a=\pm\sqrt{7}\]

Answer 12

thanks

Answer 13

yw