so by using the identities we have:

4=cos(a)cosh(b)-i sin(a)sinh(b)

so by equating real and imaginary parts we have:

(1) cos(a)cosh(b)=4 and (2) sin(a)sinh(b)=0

because b form a+i b on the left hand side is zero and a is 4. From (2) we can see that either sin(a) is zero Þ a=0 or sinh(b)=0 Þ b=0. So let's assume that b is zero, so from (1) we have:

cos(a)cosh(0)=4Þ cos(a)=4

Now we have a problem because a is a real number and cos(a) is not greater than 1 for real a, so b cannot be zero and a is zero:

cos(0)cosh(b)=4Þ cosh(b)=4Þ b=±arcosh(4)

The ± because cosh is an even function.

But a can also be 2kp for any integer k because cos(2kp) is still 1 and sin(2kp) is still 0. So:

arccos(4)=2kp±i arcosh(4).